COMPLEX HADAMARD MATRICES and APPLICATIONS Introduction 1

COMPLEX HADAMARD MATRICES AND APPLICATIONS

TEO BANICA

Abstract. A complex Hadamard matrix is a square matrix H ∈ MN (C) whose entries are on the unit circle, |Hij| = 1, and whose rows and pairwise orthogonal. The main ij 2πi/N example is the Fourier matrix, FN = (w ) with w = e . We discuss here the basic theory of such matrices, with emphasis on geometric and analytic aspects.

CONTENTS

Introduction 1 1. Hadamard matrices 5 2. Complex matrices 21 3. Roots of unity 37 4. Geometry, defect 53 5. Special matrices 69 6. Circulant matrices 85 7. Bistochastic matrices 101 8. Glow computations 117 9. Norm maximizers 133 10. Quantum groups 149 11. Subfactor theory 165 12. Fourier models 181 References 197

Introduction

A complex Hadamard matrix is a square matrix H ∈ MN (C) whose entries belong to the unit circle in the complex plane, Hij ∈ T, and whose rows are pairwise orthogonal N P with respect to the usual scalar product of C , given by < x, y >= i xiy¯i.

2010 Mathematics Subject Classiﬁcation. 15B10. Key words and phrases. Hadamard matrix, Fourier matrix. 1 2 TEO BANICA √ The orthogonality condition tells us that the rescaled matrix U = H/ N must be unitary. Thus, these matrices form a real algebraic manifold, given by: √ XN = MN (T) ∩ NUN ij 2πi/N The basic example is the Fourier matrix, FN = (w ) with w = e . In standard matrix form, and with indices i, j = 0, 1,...,N − 1, this matrix is as follows: 1 1 1 ... 1  1 w w2 . . . wN−1    1 w2 w4 . . . w2(N−1) FN =   . . . .  . . . .  1 wN−1 w(2N−1) . . . w(N−1)2 More generally, we have as example the Fourier coupling of any ﬁnite abelian group G, regarded via the isomorphism G ' Gb as a square matrix, FG ∈ MG(C): F =< i, j > G i∈G,j∈Gb

Observe that for the cyclic group G = ZN we obtain in this way the above standard Fourier matrix FN . In general, we obtain a tensor product of Fourier matrices FN .

There are many other examples of such matrices, for the most coming from various combinatorial constructions, basically involving design theory, and roots of unity. In addition, there are several deformation procedures for such matrices, leading to some more complicated constructions as well, or real algebraic geometry ﬂavor.

In general, the complex Hadamard matrices can be thought of as being “generalized Fourier matrices”, of somewhat exotic type. Due to their generalized Fourier nature, these matrices appear in a wide array of questions in mathematics and physics:

1. Operator algebras. One important concept in the theory of von Neumann algebras is that of maximal abelian subalgebra (MASA). In the ﬁnite case, where the algebra has a trace, one can talk about pairs of orthogonal MASA. In the simplest case, of the matrix ∗ algebra MN (C), the orthogonal MASA are, up to conjugation, A = ∆,B = H∆H , where ∆ ⊂ MN (C) are the diagonal matrices, and H ∈ MN (C) is Hadamard.

2. Subfactor theory. Along the same lines, but at a more advanced level, associated ∗ to any Hadamard matrix H ∈ MN (C) is the square diagram C ⊂ ∆,H∆H ⊂ MN (C) formed by the associated MASA, which is a commuting square in the sense of subfactor theory. The Jones basic construction produces, out of this diagram, an index N subfactor of the Murray-von Neumann factor R, whose computation a key problem. HADAMARD MATRICES 3

3. Quantum groups. Associated to any complex Hadamard matrix H ∈ MN (C) is a + certain quantum permutation group G ⊂ SN , obtained by factorizing the ﬂat representa- + tion π : C(SN ) → MN (C) associated to H. As a basic example here, the Fourier matrix FG produces in this way the group G itself. In general, the above-mentioned subfactor can be recovered from G, whose computation is a key problem.

4. Lattice models. According to the work of Jones, the combinatorics of the subfactor associated to an Hadamard matrix H ∈ MN (C), which by the above can be recovered + from the representation theory of the associated quantum permutation group G ⊂ SN , can be thought of as being the combinatorics of a “spin model”, in the context of link invariants, or of statistical mechanics, in an abstract, mathematical sense.

From a more applied point of view, the Hadamard matrices can be used in order to construct mutually unbiased bases (MUB) and other useful objects, which can help in connection with quantum information theory, and other quantum physics questions.

All this is quite recent, basically going back to the 00s. Regarding the known facts about the Hadamard matrices, most of them are in fact of purely mathematical nature. There are indeed many techniques that can be applied, leading to various results:

1. Algebra. In the real case, H ∈ MN (±1), the study of such matrices goes back to the beginning of the 20th century, and is quite advanced. The main problems, however, namely the Hadamard conjecture (HC) and the circulant Hadamard conjecture (CHC) are not solved yet, with no eﬃcient idea of approach in sight. Part of the real matrix techniques apply quite well to the root of unity case, H ∈ MN (Zs), with s < ∞.

2. Geometry. As already explained above,√ the N × N complex Hadamard matrices form a real algebraic manifold, XN = MN (T) ∩ NUN . This manifold is highly singular, but several interesting geometric results about it have been obtained, notably about the general structure of the singularity at a given point H ∈ XN , about the neighborhood of the Fourier matrices FG, and about the various isolated points as well.

3. Analysis. One interesting point of view on the Hadamard matrices, real√ or complex, comes from the fact that these are precisely the rescaled versions, H = NU, of the P matrices which maximize the 1-norm ||U||1 = ij |Uij| on ON ,UN respectively. When looking more generally at the local maximizers of the 1-norm, one is led into a notion of “almost Hadamard matrices”, having interesting algebraic and analytic aspects.

4. Probability. Another speculative approach, this time probabilistic, is by playing a Gale-Berlekamp type game with the matrix, in the hope that the invariants which are obtained in this way are related to the various geometric and quantum algebraic invariants, 4 TEO BANICA which are hard to compute. All this is related to the subtle fact that any unitary matrix, and so any complex Hadamard matrix as well, can be put in bistochastic form. Our aim here is to survey this material, theory and applications. Organizing all this is not easy, and we have chosen an algebra/geometry/analysis/physics lineup for our presentation, vaguely coming from the amount of background which is needed. The present text is organized in 4 parts, as follows:

(1) Sections 1-3 contain basic deﬁnitions and various algebraic results. (2) Sections 4-6 deal with diﬀerential and algebraic geometric aspects. (3) Sections 7-9 are concerned with various analytic considerations. (4) Sections 10-12 deal with various mathematical physics aspects.

There are of course many aspects of the theory which are missing from our presentation, but we will provide of course some information here, comments and references.

Acknowledgements. I would like to thank Vaughan Jones for suggesting me, back to a discussion that we had in 1997, when we ﬁrst met, to look at vertex models, and related topics. Stepping into bare Hadamard matrices is quite an experience, and very inspiring was the work of Uﬀe Haagerup on the subject, and his papers [32], [51], [52]. The present text is heavily based on a number of research papers on the subject that I wrote or co-signed, mostly during 2005–2015, and I would like to thank my cowork- ers Julien Bichon, BenoˆıtCollins, Ion Nechita, Remus Nicoar˘a,Duygu Ozteke,¨ Lorenzo Pittau, Jean-Marc Schlenker, Adam Skalski and Karol Zyczkowski.˙ Finally, many thanks go to my cats, for advice with hunting techniques, martial arts, and more. When doing linear algebra, all this knowledge is very useful. HADAMARD MATRICES 5

1. Hadamard matrices We are interested here in the complex Hadamard matrices, but we will start with some beautiful pure mathematics, regarding the real case. The deﬁnition that we need, going back to 19th century work of Sylvester [85], is as follows: Deﬁnition 1.1. An Hadamard matrix is a square binary matrix,

H ∈ MN (±1) whose rows are pairwise orthogonal, with respect to the scalar product on RN . As a first observation, we do not really need real numbers in order to talk about the Hadamard matrices, because the orthogonality condition tells us that, when comparing two rows, the number of matchings should equal the number of mismatchings. Thus, we can replace if we want the 1, −1 entries of our matrix by any two symbols, of our choice. Here is an example of an Hadamard matrix, with this convention: ♥ ♥ ♥ ♥ ♥ ♣ ♥ ♣ ♥ ♥ ♣ ♣ ♥ ♣ ♣ ♥ However, it is probably better to run away from this, and use real numbers instead, as in Definition 1.1, with the idea in mind of connecting the Hadamard matrices to the foundations of modern mathematics, namely Calculus 1 and Calculus 2. So, getting back now to the real numbers, here is a first result: Proposition 1.2. The set of the N × N Hadamard matrices is √ YN = MN (±1) ∩ NON where ON is the orthogonal group, the intersection being taken inside MN (R). √ Proof. Let H ∈ MN (±1). Since the rows of the rescaled matrix U = H/ N have norm 1, with respect to the usual scalar product on RN , we conclude that H is Hadamard precisely when U belongs to the orthogonal group ON , and so when H ∈ YN , as claimed. As an interesting consequence of the above result, which is not exactly obvious when using the design theory approach, we have the following result:

Proposition 1.3. Let H ∈ MN (±1) be an Hadamard matrix. (1) The columns of H must be pairwise orthogonal. t (2) The transpose matrix H ∈ MN (±1) is Hadamard as well.

Proof. Since the orthogonal group ON is stable under transposition, so is the set YN constructed in Proposition 1.2, and this gives both the assertions. 6 TEO BANICA

Let us study now the examples. There are many such matrices, and in order to cut a bit from the complexity, we can use the following notions:

Deﬁnition 1.4. Two Hadamard matrices are called equivalent, and we write H ∼ K, when it is possible to pass from H to K via the following operations: (1) Permuting the rows, or the columns. (2) Multiplying the rows or columns by −1. Also, we say that H is dephased when its ﬁrst row and column consist of 1 entries.

Observe that we do not include the transposition operation H → Ht in our list of allowed operations. This is because Proposition 1.3 above, while looking quite elementary, rests however on a deep linear algebra fact, namely that the transpose of an orthogonal matrix is orthogonal as well, and this can produce complications later on. Regarding the equivalence, there is of course a certain group G acting there, made of N two copies of SN , one for the rows and one for the columns, and of two copies of Z2 , once again one for the rows, and one for the columns. The equivalence classes of the Hadamard matrices are then the orbits of the action G y YN . It is possible to be a bit more explicit here, with a formula for G and so on, but we will not need this. As for the dephasing, here the terminology comes from physics, or rather from the complex Hadamard matrices. Indeed, when regarding H ∈ MN (±1) as a complex matrix, H ∈ MN (T), the −1 entries have “phases”, equal to π, and assuming that H is dephased means to assume that we have no phases, on the first row and the first column. Observe that, up to the equivalence relation, any Hadamard matrix H ∈ MN (±1) can be put in dephased form. Moreover, the dephasing operation is unique, if we use only the operations (2) in Definition 1.4, namely row and column multiplications by −1. With these notions in hand, we can formulate our first classification result:

Proposition 1.5. There is only one Hadamard matrix at N = 2, namely

1 1 W = 2 1 −1 up to the above equivalence relation for such matrices.

Proof. The matrix in the statement W2, called Walsh matrix, is clearly Hadamard. Con- versely, given H ∈ MN (±1) Hadamard, we can dephase it, as follows:

a b 1 1 1 1 → → c d ac bd 1 abcd

Now since the dephasing operation preserves the class of the Hadamard matrices, we must have abcd = −1, and so we obtain by dephasing the matrix W2. HADAMARD MATRICES 7

At N = 3 we cannot have examples, due to the orthogonality condition, which forces N to be even. At N = 4 now, we have several examples, as for instance: 1 1 1 1  1 −1 1 −1 W4 =   1 1 −1 −1 1 −1 −1 1 This matrix is a particular case of the following construction:

Proposition 1.6. If H ∈ MM (±1) and K ∈ MN (±1) are Hadamard matrices, then so is their tensor product, constructed in double index notation as follows:

H ⊗ K ∈ MMN (±1) , (H ⊗ K)ia,jb = HijKab ⊗n n In particular the Walsh matrices, WN = W2 with N = 2 , are all Hadamard.

Proof. The matrix in the statement H ⊗ K has indeed ±1 entries, and its rows Ria are pairwise orthogonal, as shown by the following computation: X < Ria,Rkc > = HijKab · HkjKcb jb X X = HijHkj KabKcb j b

= MNδikδac

As for the second assertion, this follows from this, W2 being Hadamard. Before going further, we should perhaps clarify a bit our tensor product notations. In order to write H ∈ MN (±1) the indices of H must belong to {1,...,N}, or at least to an ordered set {I1,...,IN }. But with double indices we are indeed in this latter situation, because we can use the lexicographic order on these indices. To be more precise, by using the lexicographic order on the double indices, we have the following formula:

 H11K...H1M K  . . H ⊗ K =  . .  HM1K...HMM K

As an example, by tensoring W2 with itself, we obtain the above matrix W4. Getting back now to our classiﬁcation work, here is the result at N = 4: Proposition 1.7. There is only one Hadamard matrix at N = 4, namely

W4 = W2 ⊗ W2 up to the standard equivalence relation for such matrices. 8 TEO BANICA

Proof. Consider an Hadamard matrix H ∈ M4(±1), assumed to be dephased: 1 1 1 1 1 a b c H =   1 d e f 1 g h i By orthogonality of the ﬁrst 2 rows we must have {a, b, c} = {−1, −1, 1}, and so by permuting the last 3 columns, we can further assume that our matrix is as follows: 1 1 1 1  1 −1 1 −1 H =   1 m n o  1 p q r By orthogonality of the ﬁrst 2 columns we must have {m, p} = {−1, 1}, and so by permuting the last 2 rows, we can further assume that our matrix is as follows: 1 1 1 1  1 −1 1 −1 H =   1 1 x y  1 −1 z t But this gives the result, because from the orthogonality of the rows we obtain x = y = −1, and then, with these values of x, y plugged in, from the orthogonality of the columns we obtain z = −1, t = 1. Thus, up to equivalence we have H = W4, as claimed. The case N = 5 is excluded, because the orthogonality condition forces N ∈ 2N. The point now is that the case N = 6 is excluded as well, because we have:

Proposition 1.8. The size of an Hadamard matrix must be N ∈ {2} ∪ 4N. Proof. By permuting the rows and columns or by multiplying them by −1, as to rearrange the ﬁrst 3 rows, we can always assume that our matrix looks as follows: 1 ...... 1 1 ...... 1 1 ...... 1 1 ...... 1  1 ...... 1 1 ...... 1 −1 ... − 1 −1 ... − 1   H = 1 ...... 1 −1 ... − 1 1 ...... 1 −1 ... − 1 ......  | {z } | {z } | {z } | {z } x y z t Now if we denote by x, y, z, t the sizes of the 4 block columns, as indicated, the orthogonality conditions between the ﬁrst 3 rows give the following system of equations: (1 ⊥ 2) : x + y = z + t (1 ⊥ 3) : x + z = y + t (2 ⊥ 3) : x + t = y + z HADAMARD MATRICES 9

The solution of this system being x = y = z = t, we conclude that the size of our matrix N = x + y + z + t must be a multiple of 4, as claimed. As a consequence, we are led to the study of the Hamadard matrices at: N = 8, 12, 16, 20, 24,... This study can be done either abstractly, via various algebraic methods, or with a computer, and this leads to the conclusion that the number of Hadamard matrices of size N ∈ 4N grows with N, and this in a rather exponential fashion. In particular, we are led in this way into the following statement: Conjecture 1.9 (Hadamard Conjecture (HC)). There is at least one Hadamard matrix

H ∈ MN (±1) for any integer N ∈ 4N. This conjecture, going back to the 19th century, is probably one of the most beautiful statements in combinatorics, linear algebra, and mathematics in general. Quite remark- ably, the numeric veriﬁcation so far goes up to the number of the beast: N = 666 Our purpose now will be that of gathering some evidence for this conjecture. At N = 8 we have the Walsh matrix W8. Thus, the next existence problem comes at N = 12. And here, we can use the following key construction, due to Paley [75]:

r Theorem 1.10. Let q = p be an odd prime power, deﬁne χ : Fq → {−1, 0, 1} by χ(0) = 0, χ(a) = 1 if a = b2 for some b 6= 0, and χ(a) = −1 otherwise, and ﬁnally set Qab = χ(a − b). We have then constructions of Hadamard matrices, as follows: (1) Paley 1: if q = 3(4) we have a matrix of size N = q + 1, as follows:  0 1 ... 1 −1 1   P = 1 +  .  N  . Q  −1 (2) Paley 2: if q = 1(4) we have a matrix of size N = 2q + 2, as follows: 0 1 ... 1 1 2   1 −1 1 1 P = .  : 0 → , ±1 → ± N . Q  −1 −1 1 −1 1 These matrices are skew-symmetric (H + Ht = 2), respectively symmetric (H = Ht). 10 TEO BANICA

Proof. In order to simplify the presentation, we will denote by 1 all the identity matrices, of any size, and by I all the rectangular all-one matrices, of any size as well. It is elementary to check that the matrix Qab = χ(a − b) has the following properties: t QQ = q1 − I ,QI = IQ = 0 In addition, we have the following formulae, which are elementary as well, coming from the fact that −1 is a square in Fq precisely when q = 1(4): q = 1(4) =⇒ Q = Qt q = 3(4) =⇒ Q = −Qt With these observations in hand, the proof goes as follows: (1) With our conventions for the symbols 1 and I, explained above, the matrix in the statement is as follows: 1 P 1 = I N −I 1 + Q With this formula in hand, the Hadamard matrix condition follows from: 1 1 − P 1 (P 1 )t = I I N N −I 1 + Q I 1 − Q N 0 = 0 I + 1 − Q2 N 0 = 0 N (2) If we denote by G, F the matrices in the statement, which replace respectively the 0, 1 entries, then we have the following formula for our matrix: 0 P 2 = I ⊗ F + 1 ⊗ G N I Q With this formula in hand, the Hadamard matrix condition follows from: 0 2 1 0 0 (P 2 )2 = I ⊗ F 2 + ⊗ G2 + I ⊗ (FG + GF ) N I Q 0 1 I Q q 0 1 0 0 = ⊗ 2 + ⊗ 2 + I ⊗ 0 0 q 0 1 I Q N 0 = 0 N t Finally, the last assertion is clear, from the above formulae relating Q, Q . These constructions allow us to get well beyond the Walsh matrix level, and we have the following result: HADAMARD MATRICES 11

Theorem 1.11. The HC is veriﬁed at least up to N = 88, as follows: (1) At N = 4, 8, 16, 32, 64 we have Walsh matrices. (2) At N = 12, 20, 24, 28, 44, 48, 60, 68, 72, 80, 84, 88 we have Paley 1 matrices. (3) At N = 36, 52, 76 we have Paley 2 matrices. (4) At N = 40, 56 we have Paley 1 matrices tensored with W2. However, at N = 92 these constructions (Walsh, Paley, tensoring) don’t work. Proof. First of all, the numbers in (1-4) are indeed all the multiples of 4, up to 88. As for the various assertions, the proof here goes as follows: (1) This is clear. (2) Since N − 1 takes the values q = 11, 19, 23, 27, 43, 47, 59, 67, 71, 79, 83, 87, all prime powers, we can indeed apply the Paley 1 construction, in all these cases. (3) Since N = 4(8) here, and N/2 − 1 takes the values q = 17, 25, 37, all prime powers, we can indeed apply the Paley 2 construction, in these cases. 1 1 (4) At N = 40 we have indeed P20 ⊗ W2, and at N = 56 we have P28 ⊗ W2. Finally, we have 92 − 1 = 7 × 13, so the Paley 1 construction does not work, and 92/2 = 46, so the Paley 2 construction, or tensoring with W2, does not work either. At N = 92 the situation is considerably more complicated, and we have:

Theorem 1.12. Assuming that A, B, C, D ∈ MK (±1) are circulant, symmetric, pairwise commute and satisfy A2 + B2 + C2 + D2 = 4K, the following 4K × 4K matrix  ABCD  −BA −DC  H =   −CDA −B −D −CBA is Hadamard, called of Williamson type. Moreover, such a matrix exists at K = 23.

Proof. The matrix H can be written as follows, where 1, i, j, k ∈ M4(0, 1), called the quaternion units, are the matrices describing the positions of the A, B, C, D entries: H = A ⊗ 1 + B ⊗ i + C ⊗ j + D ⊗ k Assuming now that A, B, C, D are symmetric, we have: HHt = (A ⊗ 1 + B ⊗ i + C ⊗ j + D ⊗ k)(A ⊗ 1 − B ⊗ i − C ⊗ j − D ⊗ k) = (A2 + B2 + C2 + D2) ⊗ 1 − ([A, B] − [C,D]) ⊗ i −([A, C] − [B,D]) ⊗ j − ([A, D] − [B,C]) ⊗ k Thus, if we further assume that A, B, C, D pairwise commute, and satisfy the condition A2 + B2 + C2 + D2 = 4K, we obtain indeed an Hadamard matrix. In general, ﬁnding such matrices is a diﬃcult task, and this is where Williamson’s extra assumption that A, B, C, D should be taken circulant comes from. 12 TEO BANICA

Regarding now the K = 23 construction, which produces an Hadamard matrix of order N = 92, this comes via a computer search. We refer here to [24], [97]. Things get even worse at higher values of N, where more and more complicated constructions are needed. The whole subject is quite technical, and, as already mentioned, human knowledge here stops so far at N = 666. See [1], [41], [43], [56], [64], [84]. As a conceptual ﬁnding on this subject, however, we have the recent theory of the cocyclic Hadamard matrices. The basic deﬁnition here is as follows:

Deﬁnition 1.13. A cocycle on a ﬁnite group G is a matrix H ∈ MG(±1) satisfying:

H11 = 1 ,HghHgh,k = Hg,hkHhk If the rows of H are pairwise orthogonal, we say that H is a cocyclic Hadamard matrix. Here the deﬁnition of the cocycles is the usual one, with the equations coming from the fact that F = Z2 × G must be a group, with multiplication as follows:

(u, g)(v, h) = (Hgh · uv, gh)

As a basic example, the Walsh matrix H = W2n is cocyclic, coming from the group n G = Z2 , with cocycle Hgh = (−1) . As explained in [42], many other known examples of Hadamard matrices are cocyclic, and this leads to the following conjecture: Conjecture 1.14 (Cocyclic Hadamard Conjecture). There is at least one cocyclic Hada- mard matrix H ∈ MN (±1), for any N ∈ 4N. Having such a statement formulated is certainly a big advance with respect to the HC, and this is probably the main achievement of modern Hadamard matrix theory. However, in what regards a potential proof, there is no serious idea here, at least so far.

One potential way of getting away from these questions is that of looking at various special classes of Hadamard matrices. However, this is not really the case, because passed a few trivialities, the existence of special Hadamard matrices is generally subject to an improvement of the HC, as in the cocyclic case, or to diﬃcult non-existence questions. Illustrating and quite famous here is the situation in the circulant case. Given a vector N γ ∈ (±1) , one can ask whether the matrix H ∈ MN (±1) deﬁned by Hij = γj−i is Hadamard or not. Here is a solution to the problem: −1 1 1 1   1 −1 1 1  K4 =    1 1 −1 1  1 1 1 −1

4 P More generally, any vector γ ∈ (±1) satisfying γi = ±1 is a solution to the problem. The following conjecture, going back to [84], states that there are no other solutions: HADAMARD MATRICES 13

Conjecture 1.15 (Circulant Hadamard Conjecture (CHC)). There is no circulant Ha- damard matrix of size N × N, for any N 6= 4. The fact that such a simple-looking problem is still open might seem quite surprising. Indeed, if we denote by S ⊂ {1,...,N} the set of positions of the −1 entries of γ, the Hadamard matrix condition is simply |S ∩ (S + k)| = |S| − N/4, for any k 6= 0, taken modulo N. Thus, the above conjecture simply states that at N 6= 4, such a set S cannot exist. Let us record here this latter statement, originally due to Ryser [83]: Conjecture 1.16 (Ryser Conjecture). Given an integer N > 4, there is no set S ⊂ {1,...,N} satisfying the condition |S ∩ (S + k)| = |S| − N/4 for any k 6= 0, taken modulo N. There has been a lot of work on this conjecture, starting with [83]. However, as it was the case with the HC, all this leads to complicated combinatorics, design theory, algebra and number theory, and so on, and there is no serious idea here, at least so far.

All this might seem a bit depressing, but there are at least two potential exits from the combinatorial and algebraic theory of Hadamard matrices, namely: (1) Do analysis. There are many things that can be done here, starting with the Hadamard determinant bound [53], and we will discuss this below. Whether all this can help or not in relation with the HC and CHC remains to be seen, but at least we’ll have some fun, and do some interesting mathematics. (2) Do physics. When allowing the entries of H to be complex numbers, both the HC ij 2πi/N and the CHC dissapear, because the Fourier matrix FN = (w ) with w = e is an example of such matrix at any N ∈ N, which in addition can be put in circulant form. We will discuss this later, starting from section 2 below. So, let us step now into analytic questions. The ﬁrst result here, found in 1893 by Hadamard [53], about 25 years after Sylvester’s 1867 founding paper [85], and which actually led to such matrices being called Hadamard, is as follows:

Theorem 1.17. Given a matrix H ∈ MN (±1), we have | det(H)| ≤ N N/2 with equality precisely when H is Hadamard. Proof. We use here the fact, which often tends to be forgotten, that the determinant of a system of N vectors in RN is the signed volume of the associated parallelepiped:

det(H1,...,HN ) = ±vol < H1,...,HN > This is actually the deﬁnition of the determinant, in case you have forgotten the basics (!), with the need for the sign coming for having good additivity properties. 14 TEO BANICA

In the case where our vectors take their entries in ±1, we therefore have the following inequality, with equality precisely when our vectors are pairwise orthogonal: √ N | det(H1,...,HN )| ≤ ||H1|| × ... × ||HN || = ( N) Thus, we have obtained the result, straight from the deﬁnition of det. The above result is quite interesting, philosophically speaking. Let us recall indeed from Proposition 1.2 that the set formed by the N × N Hadamard matrices is: √ YN = MN (±1) ∩ NON

Thus, what we have in Theorem 1.17 is an analytic method for locating YN inside MN (±1). This suggests doing many geometric and analytic things, as for instance looking at the maximizers of | det(H)| at values N ∈ N which are not multiples of 4. These latter matrices are called “quasi-Hadamard”, and we refer here to [76]. √ From a “dual” point of view, the question of locating YN inside NON , once again via analytic methods, makes sense as well. The result here, from [14], is as follows:

Theorem 1.18. Given a matrix U ∈ ON we have √ ||U||1 ≤ N N √ with equality precisely when H = U/ N is Hadamard.

Proof. We have indeed the following estimate, valid for any U ∈ ON : !1/2 √ X X 2 ||U||1 = |Uij| ≤ N |Uij| = N N ij ij √ The equality√ case holds when |Uij| = N for any i, j, which amounts in saying that H = U/ N must satisfy H ∈ MN (±1), and so that H must be Hadamard. As a ﬁrst comment here, the above Cauchy-Schwarz estimate can be improved with a H¨olderestimate, the conclusion being that the rescaled Hadamard matrices maximize the p-norm on ON at any p ∈ [1, 2), and minimize it at any p ∈ (2, ∞]. We will discuss this in section 9 below, with full details, directly in the complex case. As it was the case with the Hadamard determinant bound, all this suggests doing some further geometry and analysis, this time on the Lie group ON , notably with a notion of “almost Hadamard matrix” at stake. We will be back to this in section 9 below.

Let us discuss now, once again as an introduction to analytic topics, yet another such result. We recall that a matrix H ∈ MN (R) is called row-stochastic when the sums on the rows are all equal, column-stochastic when the same is true for columns, and bistochastic when this is true for both rows and columns, the common sum being the same. With this terminology, we have the following well-known result: HADAMARD MATRICES 15

Theorem 1.19. For an Hadamard matrix H ∈ MN (±1), the excess, X E(H) = Hij ij √ satisﬁes |E(H)| ≤ N N, with equality if and only if H is bistochastic.

N Proof. In terms of the all-one vector ξ = (1)i ∈ R , we have: X X X E(H) = Hij = Hijξjξi = (Hξ)iξi =< Hξ, ξ > ij ij i √ Now by using the Cauchy-Schwarz inequality, along with the fact that U = H/ N is orthogonal, and hence of norm 1, we obtain, as claimed: √ |E(H)| ≤ ||Hξ|| · ||ξ|| ≤ ||H|| · ||ξ||2 = N N

Regarding now the equality case, this requires the vectors√Hξ, ξ to be proportional, and so our matrix H to be row-stochastic. But since U = H/ N is orthogonal, Hξ ∼ ξ t is equivalent to H ξ ∼ ξ, and so H must be bistochastic, as claimed. There are many known interesting results on the bistochastic Hadamard matrices, and we will be back to this in section 7 below, directly in the complex setting.

One interesting question, that we would like to discuss now, is that of computing the law of the excess over the equivalence class of H. Let us start with:

Deﬁnition 1.20. The glow of H ∈ MN (±1) is the probability measure µ ∈ P(Z) describ- P ing the distribution of the excess, E = ij Hij, over the equivalence class of H. Since the excess is invariant under permutations of rows and columns, we can restrict the attention to the matrices He ' H obtained by switching signs on rows and columns. N N More precisely, let (a, b) ∈ Z2 × Z2 , and consider the following matrix:

Heij = aibjHij

N N We can regard the sum of entries of He as a random variable, over the group Z2 × Z2 , and we have the following equivalent description of the glow:

N N Proposition 1.21. Given a matrix H ∈ MN (±1), if we deﬁne ϕ : Z2 × Z2 → Z by X ϕ(a, b) = aibjHij ij then the glow µ is the probability measure on Z given by µ({k}) = P (ϕ = k). 16 TEO BANICA

Proof. The function ϕ in the statement can indeed be regarded as a random variable over N N the group Z2 × Z2 , with this latter group being endowed with its uniform probability measure P . The distribution µ of this variable ϕ is then given by: 1 n o µ({k}) = # (a, b) ∈ N × N ϕ(a, b) = k 4N Z2 Z2 By the above discussion, this distribution is exactly the glow. The terminology in Deﬁnition 1.20 comes from the following picture. Assume that we have a square city, with N horizontal streets and N vertical streets, and with street lights at each crossroads. When evening comes the lights are switched on at the positions (i, j) where Hij = 1, and then, all night long, they are randomly switched on and oﬀ, with the help of 2N master switches, one at the end of each street: → ♦ ♦ ♦ ♦ → ♦ × ♦ × → ♦ ♦ × × → ♦ × × ♦

↑ ↑ ↑ ↑ With this picture in mind, µ describes indeed the glow of the city. At a more advanced level now, all this is related to the Gale-Berlekamp game [50], [82], and this is where our main motivation for studying it comes from. In order to compute the glow, it is useful to have in mind the following picture:

b1 . . . bN ↓ ↓ (a1) → H11 ...H1N ⇒ S1 ...... (aN ) → HN1 ...HNN ⇒ SN Here the columns of H have been multiplied by the entries of the horizontal switching vector b, the resulting sums on rows are denoted S1,...,SN , and the vertical switching vector a still has to act on these sums, and produce the glow component at b. With this picture in mind, we ﬁrst have the following result, from [9]:

Proposition 1.22. The glow of a matrix H ∈ MN (±1) is given by 1 X µ = β (c ) ∗ ... ∗ β (c ) 2N 1 1 N N N b∈Z2 ∗c δr+δ−r where βr(c) = 2 , and cr = #{r ∈ |S1|,..., |SN |}, with S = Hb. HADAMARD MATRICES 17

Proof. We use the interpretation of the glow explained above. So, consider the decomposition of the glow over b components: 1 X µ = µ 2N b N b∈Z2

With the notation S = Hb, the numbers S1,...,SN are the sums on the rows of the matrix Heij = Hijaibj. Thus the glow components are given by:

µb = law (±S1 ± S2 ... ± SN ) By permuting now the sums on the right, we have the following formula: µb = law ±0 ... ± 0 ±1 ... ± 1 ...... ±N... ± N | {z } | {z } | {z } c0 c1 cN Now since the ± variables each follow a Bernoulli law, and these Bernoulli laws are independent, we obtain a convolution product as in the statement. We will need the following elementary fact:

Proposition 1.23. Let H ∈ MN (±1) be an Hadamard matrix of order N ≥ 4.

(1) The sums of entries on rows S1,...,SN are even, and equal modulo 4. (2) If the sums on the rows S1,...,SN are all 0 modulo 4, then the number of rows whose sum is 4 modulo 8 is odd for N = 4(8), and even for N = 0(8). Proof. (1) Let us pick two rows of our matrix, and then permute the columns such that these two rows look as follows: 1 ...... 1 1 ...... 1 −1 ... − 1 −1 ... − 1 1 ...... 1 −1 ... − 1 1 ...... 1 −1 ... − 1 | {z } | {z } | {z } | {z } a b c d N We have a + b + c + d = N, and by orthogonality a + d = b + c, so a + d = b + c = 2 . Now since N/2 is even, we conclude that b = c(2), and this gives the result. (2) In the case where H is “row-dephased”, in the sense that its ﬁrst row consists of 1 entries only, the row sums are N, 0,..., 0, and so the result holds. In general now, by permuting the columns we can assume that our matrix looks as follows: 1 ...... 1 −1 ... − 1 . . H =  . .  |{z} |{z} x y

We have x + y = N = 0(4), and since the first row sum S1 = x − y is by assumption 0 modulo 4, we conclude that x, y are even. In particular, since y is even, the passage from H to its row-dephased version He can be done via y/2 double sign switches. Now, in view of the above, it is enough to prove that the conclusion in the statement is stable under a double sign switch. So, let H ∈ MN (±1) be Hadamard, and let us perform 18 TEO BANICA to it a double sign switch, say on the first two columns. Depending on the values of the entries on these first two columns, the total sums on the rows change as follows: + + ...... : S → S − 4 + − ...... : S → S − + ...... : S → S − − ...... : S → S + 4 We can see that the changes modulo 8 of the row sum S occur precisely in the first and in the fourth case. But, since the first two columns of our matrix H ∈ MN (±1) are orthogonal, the total number of these cases is even, and this finishes the proof. Observe that Proposition 1.22 and Proposition 1.23 (1) show that the glow of an Ha- damard matrix of order N ≥ 4 is supported by 4Z. With this in hand, we have:

Theorem 1.24. Let H ∈ MN (±1) be an Hadamard matrix of order N ≥ 4, and denote by µeven, µodd the mass one-rescaled restrictions of µ ∈ P(4Z) to 8Z, 8Z + 4. 3 even 1 odd (1) At N = 0(8) we have µ = 4 µ + 4 µ . 1 even 3 odd (2) At N = 4(8) we have µ = 4 µ + 4 µ . Proof. We use the glow decomposition over b components, from Proposition 1.22: 1 X µ = µ 2N b N b∈Z2 The idea is that the decomposition formula in the statement will occur over averages N−1 of the following type, over truncated sign vectors c ∈ Z2 : 1 µ0 = (µ + µ ) c 2 +c −c Indeed, we know from Proposition 1.23 (1) that modulo 4, the sums on rows are either 0,..., 0 or 2,..., 2. Now since these two cases are complementary when pairing switch vectors (+c, −c), we can assume that we are in the case 0,..., 0 modulo 4. Now by looking at this sequence modulo 8, and letting x be the number of 4 components, so that the number of 0 components is N − x, we have:   1 1 (µ+c + µ−c) = law(±0 ... ± 0 ±4 ... ± 4) + law(±2 ... ± 2) 2 2 | {z } | {z } | {z } N−x x N Now by using Proposition 1.23 (2), the ﬁrst summand splits 1−0 or 0−1 on 8Z, 8Z+4, depending on the class of N modulo 8. As for the second summand, since N is even this 1 1 3 1 always splits 2 − 2 on 8Z, 8Z + 4. So, by making the average we obtain either a 4 − 4 or 1 3 a 4 − 4 splitting on 8Z, 8Z + 4, depending on the class of N modulo 8, as claimed. HADAMARD MATRICES 19

Various computer simulations suggest that the measures µeven, µodd don’t have further general algebraic properties. Analytically speaking now, we have:

Theorem 1.25. The glow moments of H ∈ MN (±1) are given by: Z E 2p = (2p)!! + O(N −1) N N N Z2 ×Z2 In particular the variable E/N becomes Gaussian in the N → ∞ limit.

Proof. Let Peven(r) ⊂ P (r) be the set of partitions of {1, . . . , r} having all blocks of even P size. The moments of the variable E = ij aibjHij are then given by: Z Z Z r X E = Hi1x1 ...Hirxr ai1 . . . air bx1 . . . bxr N N N N Z2 ×Z2 ix Z2 Z2 X X = Hi1x1 ...Hirxr

π,σ∈Peven(r) ker i=π,ker x=σ

Thus the moments decompose over partitions π ∈ Peven(r), with the contributions being obtained by integrating the following quantities: X X C(σ) = Hi1x1 ...Hirxr · ai1 . . . air ker x=σ i Now by Möbiusinversion, we obtain a formula as follows: Z X Er = K(π)N |π|I(π) N × N Z2 Z2 π∈Peven(r) Here K(π) = P µ(π, σ), where µ is the Möbiusfunction of P (r), and, with σ∈Peven(r) even N the convention that H1,...,HN ∈ Z2 are the rows of H: * + X Y 1 Y I(π) = H , 1 N ir i b∈π r∈b With this formula in hand, the first assertion follows, because the biggest elements of the lattice Peven(2p) are the (2p)!! partitions consisting of p copies of a 2-block. As for the second assertion, this follows from the moment formula, and from the fact that the glow of H ∈ MN (±1) is real, and symmetric with respect to 0. See [8]. We will be back to all this in section 8 below, in the complex matrix setting.

Finally, some interesting analytic results can be obtained by exiting the square matrix setting, and looking at the rectangular matrix case. Let us start with: 20 TEO BANICA

Deﬁnition 1.26. A partial Hadamard matrix (PHM) is a matrix

H ∈ MM×N (±1) having its rows pairwise orthogonal. These matrices are quite interesting objects, appearing in connection with various questions in combinatorics. The motivating examples are the Hadamard matrices H ∈ MN (±1), and their M × N submatrices, with M ≤ N. See [54], [58], [84], [94]. In their paper [44], de Launey and Levin were able to count these matrices, in the asymptotic limit N ∈ 4N, N → ∞. Their method is based on:

Proposition 1.27. The probability for a random H ∈ MM×N (±1) to be partial Hadamard equals the probability for a length N random walk with increments drawn from n o M E = (eie¯j)i

Proof. Indeed, with T (e) = (eie¯j)i

Theorem 1.28. The probability for a random H ∈ MM×N (±1) to be PHM is 2(M−1)2 PM ' q M (2πN)( 2 ) in the N ∈ 4N, N → ∞ limit. Proof. According to Proposition 1.27 above, we have: ( ) 1 X 1 X P = # ξ , . . . , ξ ∈ E ξ = 0 = δ M q(M−1)N 1 N i q(M−1)N Σξi,0 i ξ1,...,ξN ∈E M By using the Fourier inversion formula we have, with D = 2 : 1 Z i<λ,Σξi> δΣξi,0 = D e dλ (2π) [−π,π]D After many non-trivial computations, this leads to the result. See [44]. All this is extremely interesting. Let us mention as well that for the general matrices H ∈ MM×N (±1), which are not necessarily PHM, such statistics can be deduced from the work of Tau-Vu [91]. Finally, there is an extension of the notion of PHM in the complex case, and we will be back to this later on, in section 3 below. HADAMARD MATRICES 21

2. Complex matrices

We have seen that the Hadamard matrices H ∈ MN (±1) are very interesting combinatorial objects. In what follows, we will be interested in their complex versions: Deﬁnition 2.1. A complex Hadamard matrix is a square complex matrix

H ∈ MN (C) whose entries are on the unit circle, Hij ∈ T, and whose rows are pairwise orthogonal. Here, and in what follows, the scalar product is the usual one on CN , taken to be linear in the ﬁrst variable and antilinear in the second one: X < x, y >= xiy¯i i As basic examples of complex Hamadard matrices, we have of course the real Hadamard matrices, H ∈ MN (±1). We will see that there are many other examples. Let us start by extending some basic results from the real case. First, we have: Proposition 2.2. The set of the N × N complex Hadamard matrices is the real algebraic manifold √ XN = MN (T) ∩ NUN where UN is the unitary group, the intersection being taken inside MN (C). √ Proof. Let H ∈ MN (T). Then H is Hadamard if and only if its rescaling U = H/ N belongs to the unitary group UN , and so when H ∈ YN , as claimed.

The above manifold XN , while appearing by deﬁnition as an intersection of smooth manifolds, is very far from being smooth. We will be back to this, later on. As a basic consequence of the above result, we have:

Proposition 2.3. Let H ∈ MN (C) be an Hadamard matrix. (1) The columns of H must be pairwise orthogonal. t ¯ ∗ (2) The matrices H , H,H ∈ MN (C) are Hadamard as well.

Proof. We use the well-known fact that if a matrix is unitary, U ∈ UN , then so is its ¯ ¯ complex conjugate U = (Uij), the inversion formulae being as follows: U ∗ = U −1 ,U t = U¯ −1 t ¯ ∗ Thus the unitary group UN is stable under the operations U → U ,U → U,U → U , and it follows that the algebraic manifold XN constructed in Proposition 2.2 is stable as well under these operations. But this gives all the assertions. Let us introduce now the following equivalence notion for the complex Hadamard matrices, taking into account some basic operations which can be performed: 22 TEO BANICA

Definition 2.4. Two complex Hadamard matrices are called equivalent, and we write H ∼ K, when it is possible to pass from H to K via the following operations: (1) Permuting the rows, or permuting the columns. (2) Multiplying the rows or columns by numbers in T. Also, we say that H is dephased when its first row and column consist of 1 entries. The same remarks as in the real case apply. For instance, we have not taken into account the results in Proposition 2.3 when formulating the above definition, because the operations H → Ht, H,H¯ ∗ are far more subtle than those in (1,2) above. At the level of the examples now, we have the following basic construction, which works at any N ∈ N, in stark contrast with what happens in the real case: ij 2πi/N Theorem 2.5. The Fourier matrix, FN = (w ) with w = e , which in standard matrix form, with indices i, j = 0, 1,...,N − 1, is as follows, 1 1 1 ... 1  1 w w2 . . . wN−1    1 w2 w4 . . . w2(N−1) FN =   . . . .  . . . .  1 wN−1 w(2N−1) . . . w(N−1)2 is a complex Hadamard matrix, in dephased form. Proof. By using the standard fact that the averages of complex numbers correspond to barycenters, we conclude that the scalar products between the rows of FN are: X aj −bj X (a−b)j < Ra,Rb >= w w = w = Nδab j j

Thus FN is indeed a complex Hadamard matrix. As for the fact that FN is dephased, this follows from our convention i, j = 0, 1,...,N − 1, which is there for this. Thus, there is no analogue of the HC in the complex case. We will see later on, in section 6 below, that the Fourier matrix FN can be put in circulant form, so there is no analogue of the CHC either, in this setting. This is of course very good news. We should mention, however, that the HC and CHC do have some complex extensions, which are of technical nature, by restricting the attention to the Hadamard matrices formed by roots of unity of a given order. We will discuss this in section 3 below. As a ﬁrst classiﬁcation result now, in the complex case, we have:

Proposition 2.6. The Fourier matrices F2,F3, which are given by 1 1 1  1 1 F = ,F = 1 w w2 2 1 −1 3   1 w2 w with w = e2πi/3 are the only Hadamard matrices at N = 2, 3, up to equivalence. HADAMARD MATRICES 23

Proof. The proof at N = 2 is similar to the proof of Proposition 1.5. Regarding now the case N = 3, consider an Hadamard matrix H ∈ M3(T), in dephased form: 1 1 1 H = 1 x y 1 z t The orthogonality conditions between the rows of this matrix read: (1 ⊥ 2) : x + y = −1

(1 ⊥ 3) : z + t = −1 (2 ⊥ 3) : xz¯ + yt¯= −1 Now observe that the equation p+q = −1 with p, q ∈ T tells us that the triangle having vertices at 1, p, q must be equilateral, and so that {p, q} = {w, w2}, with w = e2πi/3. By using this fact, for the ﬁrst two equations, we conclude that we must have {x, y} = {w, w2} and {z, t} = {w, w2}. As for the third equation, this tells us that we must have x 6= z. Thus, our Hadamard matrix H is either the Fourier matrix F3, or is the matrix obtained from F3 by permuting the last two columns, and we are done. In order to deal now with the case N = 4, we already know, from our study in the real case, that we will need tensor products. So, let us formulate: Deﬁnition 2.7. The tensor product of complex Hadamard matrices is given, in double indices, by (H ⊗ K)ia,jb = HijKab. In other words, we have the formula

 H11K...H1M K  . . H ⊗ K =  . .  HM1K...HMM K by using the lexicographic order on the double indices. In order to advance, our ﬁrst task will be that of tensoring the Fourier matrices. And here, we have the following statement, reﬁning and generalizing Theorem 2.5:

Theorem 2.8. Given a ﬁnite abelian group G, with dual group Gb = {χ : G → T}, consider the Fourier coupling FG : G × Gb → T, given by (i, χ) → χ(i). (1) Via the standard isomorphism G ' Gb, this Fourier coupling can be regarded as a square matrix, FG ∈ MG(T), which is a complex Hadamard matrix. (2) In the case of the cyclic group G = ZN we obtain in this way, via the standard identiﬁcation ZN = {1,...,N}, the Fourier matrix FN .

(3) In general, when using a decomposition G = ZN1 × ... × ZNk , the corresponding

Fourier matrix is given by FG = FN1 ⊗ ... ⊗ FNk . 24 TEO BANICA

Proof. This follows indeed from some basic facts from group theory: (1) With the identiﬁcation G ' Gb made our matrix is given by (FG)iχ = χ(i), and the scalar products between the rows are computed as follows: X X < Ri,Rj >= χ(i)χ(j) = χ(i − j) = |G| · δij χ χ Thus, we obtain indeed a complex Hadamard matrix. (2) This follows from the well-known and elementary fact that, via the identiﬁcations ij 2πi/N ZN = ZcN = {1,...,N}, the Fourier coupling here is (i, j) → w , with w = e . (3) We use here the following well-known formula, for the duals of products:

H\× K = Hb × Kb At the level of the corresponding Fourier couplings, we obtain from this:

FH×K = FH ⊗ FK Now by decomposing G into cyclic groups, as in the statement, and by using (2) for the cyclic components, we obtain the formula in the statement. As a ﬁrst application of this result, we have:

n Proposition 2.9. The Walsh matrix, WN with N = 2 , which is given by 1 1 ⊗n W = N 1 −1

n is the Fourier matrix of the ﬁnite abelian group KN = Z2 .

Proof. We have indeed W2 = F2 = FK2 , and by taking tensor powers we obtain from this n that we have WN = FKN , for any N = 2 . By getting back now to our classiﬁcation work, the possible abelian groups at N = 4, that we can use, are the cyclic group Z4, which produces the Fourier matrix F4, and the Klein group K4 = Z2 × Z2, which produces the Walsh matrix W4. The point, however, is that, besides F4,W4, there are many other complex Hadamard matrices at N = 4. Indeed, we can use here the following version of the tensor product construction, coming from Dit¸˘a’spaper [47], involving parameters:

Proposition 2.10. If H ∈ MM (T) and K ∈ MN (T) are Hadamard, then so are the following two matrices, for any choice of a parameter matrix Q ∈ MM×N (T): (1) H ⊗Q K ∈ MMN (T), given by (H ⊗Q K)ia,jb = QibHijKab. (2) H Q ⊗ K ∈ MMN (T), given by (H Q ⊗ K)ia,jb = QjaHijKab. These are called right and left Dit¸˘adeformations of H ⊗ K, with parameter Q. HADAMARD MATRICES 25

Proof. The rows Ria of the matrix H ⊗Q K from (1) are indeed pairwise orthogonal:

X ¯ ¯ ¯ < Ria,Rkc > = QibHijKab · QkbHkjKcb jb X ¯ = Mδik KabKcb b = MNδikδac

As for the rows Lia of the matrix HQ ⊗ K from (2), these are orthogonal as well:

X ¯ ¯ ¯ < Lia,Lkc > = QjaHijKab · QjcHkjKcb jb X ¯ = Nδac HijHkj j

= MNδikδac

Thus, both the matrices in the statement are Hadamard, as claimed.

As a ﬁrst observation, when the parameter matrix is the all-one matrix I ∈ MM×N (T), we obtain in this way the usual tensor product of our matrices:

H ⊗I K = HI ⊗ K = H ⊗ K

As a non-trivial example now, the right deformations of the Walsh matrix W4 = F2 ⊗F2, p q with arbitrary parameter matrix Q = (r s), are given by:

p q p q  1 1 1 1 p −q p −q F2 ⊗Q F2 = ⊗  =   1 −1 p q 1 −1 r s −r −s r s r −s −r s

This follows indeed by carefully working out what happens, by using the lexicographic order on the double indices, as explained after Proposition 1.6 above. To be more precise, the usual tensor product W4 = F2 ⊗ F2 appears as follows: ia\jb 00 01 10 11       00 1 1 1 1  W4 =    01 1 −1 1 −1  10 1 1 −1 −1 11 1 −1 −1 1 26 TEO BANICA

The corresponding values of the parameters Qib to be inserted are as follows: ia\jb 00 01 10 11       00 Q00 Q01 Q00 Q01 (Qib) =    01 Q00 Q01 Q00 Q01  10 Q10 Q11 Q10 Q11 11 Q10 Q11 Q10 Q11 p q With the notation Q = (r s), this latter matrix becomes: ia\jb 00 01 10 11      00 p q p q  (Qib) =    01 p q p q   10 r s r s  11 r s r s

Now by pointwise multiplying this latter matrix with the matrix W4 given above, we obtain the announced formula for the deformed tensor product F2 ⊗Q F2. As for the left deformations of W4 = F2 ⊗ F2, once again with arbitrary parameter p q matrix Q = (r s), these are given by a similar formula, as follows: p p r r  1 1 1 1 q −q s −s F2Q ⊗ F2 =   ⊗ =   1 −1 p q 1 −1 p p −r −r r s q −q −s s

Observe that this latter matrix is transpose to F2 ⊗Q F2. However, this is something accidental, coming from the fact that F2, and so W4 as well, are self-transpose. With the above constructions in hand, we have the following result: Theorem 2.11. The only complex Hadamard matrices at N = 4 are, up to the standard equivalence relation, the matrices 1 1 1 1  s 1 −1 1 −1 F =   4 1 s −1 −s 1 −s −1 s with s ∈ T, which appear as right Dit¸˘adeformations of W4 = F2 ⊗ F2. s Proof. First of all, the matrix F4 is indeed Hadamard, appearing from the construction in Proposition 2.10, assuming that the parameter matrix there Q ∈ M2(T) is dephased: 1 1 Q = 1 s HADAMARD MATRICES 27

Observe also that, conversely, any right Dit¸˘adeformation of W4 = F2 ⊗ F2 is of this p q form. Indeed, if we consider such a deformation, with general parameter matrix Q = (r s) s0 0 as above, by dephasing we obtain an equivalence with F4 , where s = ps/qr: p q p q   1 1 1 1  p −q p −q  1 −1 1 −1    →   r s −r −s r/p s/q −r/p −s/q r −s −r s r/p −s/q −r/p s/q 1 1 1 1  1 −1 1 −1  →   1 ps/qr −1 −ps/qr 1 −ps/qr −1 ps/qr

s It remains to prove that the matrices F4 are non-equivalent, and that any complex s Hadamard matrix H ∈ M4(T) is equivalent to one of these matrices F4 . But this follows by using the same kind of arguments as in the proof of Proposition 1.7, and from the proof of Proposition 2.6. Indeed, let us ﬁrst dephase our matrix:

1 1 1 1 1 a b c H =   1 d e f 1 g h i

We use now the fact, coming from plane geometry, that the solutions x, y, z, t ∈ T of the equation x + y + z + t = 0 are given by {x, y, z, t} = {p, q, −p, −q}, with p, q ∈ T. In our case, we have 1 + a + d + g = 0, and so up to a permutation of the last 3 rows, our matrix must look at follows, for a certain s ∈ T: 1 1 1 1 1 −1 b c H =   1 s e f 1 −s h i

In the case s = ±1 we can permute the middle two columns, then repeat the same reasoning, and we end up with the matrix in the statement. In the case s 6= ±1 we have 1 + s + e + f = 0, and so −1 ∈ {e, f}. Up to a permutation of the last columns, we can assume e = −1, and our matrix becomes:

1 1 1 1  1 −1 b c  H =   1 s −1 −s 1 −s h i 28 TEO BANICA

Similarly, from 1 − s + h + i = 0 we deduce that −1 ∈ {h, i}. In the case h = −1 our matrix must look as follows, and we are led to the matrix in the statement: 1 1 1 1  1 −1 b c  H =   1 s −1 −s 1 −s −1 i As for the remaining case i = −1, here our matrix must look as follows: 1 1 1 1  1 −1 b c  H =   1 s −1 −s 1 −s h −1 We obtain from the last column c = s, then from the second row b = −s, then from the third column h = s, and so our matrix must be as follows: 1 1 1 1  1 −1 −s s  H =   1 s −1 −s 1 −s s −1 But, in order for the second and third row to be orthogonal, we must have s ∈ R, and so s = ±1, which contradicts our above assumption s 6= ±1. Thus, we are done with the proof of the main assertion. As for the fact that the matrices in the statement are indeed not equivalent, this is standard as well. See [88].

At N = 5 now, the situation is considerably more complicated, with F5 being the only known example, but with the proof of its uniqueness being highly nontrivial. The key technical result here, due to Haagerup [51], is as follows:

Proposition 2.12. Given an Hadamard matrix H ∈ M5(T), chosen dephased, 1 1 1 1 1 1 a x ∗ ∗   H = 1 y b ∗ ∗   1 ∗ ∗ ∗ ∗ 1 ∗ ∗ ∗ ∗ the numbers a, b, x, y must satisfy (x − y)(x − ab)(y − ab) = 0. Proof. This is something quite surprising, and tricky, the proof in [51] being as follows. Let us look at the upper 3-row truncation of H, which is of the following form: 1 1 1 1 1 0 H = 1 a x p q 1 y b r s HADAMARD MATRICES 29

By using the orthogonality of the rows, we have: (1 + a + x)(1 + ¯b +y ¯)(1 +ay ¯ + bx¯) = −(p + q)(r + s)(¯pr +qs ¯ ) On the other hand, by using p, q, r, s ∈ T, we have: (p + q)(r + s)(¯pr +qs ¯ ) = (r + pqs¯ +pqr ¯ + s)(¯r +s ¯) = 1 + pq¯rs¯ +pq ¯ +rs ¯ + rs¯ + pq¯ +pqr ¯ s¯ + 1 = 2Re(1 + pq¯ + rs¯ + pqr¯ s¯) = 2Re[(1 + pq¯)(1 + rs¯)] We conclude that we have the following formula, involving a, b, x, y only: (1 + a + x)(1 + ¯b +y ¯)(1 +ay ¯ + bx¯) ∈ R Now this is a product of type (1+α)(1+β)(1+γ), with the ﬁrst summand being 1, and with the last summand, namely αβγ, being real as well, as shown by the above general p, q, r, s ∈ T computation. Thus, when expanding, and we are left with: (a + x) + (¯b +y ¯) + (¯ay + bx¯) + (a + x)(¯b +y ¯) + (a + x)(¯ay + bx¯) + (¯b +y ¯)(¯ay + bx¯) ∈ R By expanding all the products, our formula looks as follows: a + x + ¯b +y ¯ +ay ¯ + bx¯ + a¯b + ay¯ + ¯bx + xy¯ + 1 + abx¯ +axy ¯ + b +a ¯¯by +x ¯ +a ¯ + bx¯y¯ ∈ R By removing from this all terms of type z +z ¯, we are left with: a¯b + xy¯ + abx¯ +a ¯¯by +axy ¯ + bx¯y¯ ∈ R Now by getting back to our Hadamard matrix, all this remains true when transposing it, which amounts in interchanging x ↔ y. Thus, we have as well: a¯b +xy ¯ + aby¯ +a ¯¯bx +axy ¯ + bx¯y¯ ∈ R By substracting now the two equations that we have, we obtain: xy¯ − xy¯ + ab(¯x − y¯) +a ¯¯b(y − x) ∈ R Now observe that this number, say Z, is purely imaginary, because Z¯ = −Z. Thus our equation reads Z = 0. On the other hand, we have the following formula: abxyZ = abx2 − aby2 + a2b2(y − x) + xy(y − x) = (y − x)(a2b2 + xy − ab(x + y)) = (y − x)(ab − x)(ab − y) Thus, our equation Z = 0 corresponds to the formula in the statement. 30 TEO BANICA

By using the above result, we are led to the following theorem, also from [51]: Theorem 2.13. The only Hadamard matrix at N = 5 is the Fourier matrix, 1 1 1 1 1  1 w w2 w3 w4  2 4 3 F5 = 1 w w w w   3 4 2 1 w w w w  1 w4 w3 w2 w with w = e2πi/5, up to the standard equivalence relation for such matrices.

Proof. Assume that have an Hadamard matrix H ∈ M5(T), chosen dephased, and written as in Proposition 2.12, with emphasis on the upper left 2 × 2 subcorner: 1 1 1 1 1 1 a x ∗ ∗   H = 1 y b ∗ ∗   1 ∗ ∗ ∗ ∗ 1 ∗ ∗ ∗ ∗ We know from Proposition 2.12, applied to H itself, and to its transpose Ht as well, that the entries a, b, x, y must satisfy the following equations: (a − b)(a − xy)(b − xy) = 0 (x − y)(x − ab)(y − ab) = 0 This is of course something very strong, and these equations are actually valid all across the matrix, by permuting rows and columns. The idea will be that by doing some combinatorics, sometimes combined with a few tricks, this will lead to the result. Our first claim is that, by doing some combinatorics, we can actually obtain from this a = b and x = y, up to the equivalence relation for the Hadamard matrices: 1 1 1 1 1 1 a x ∗ ∗   H ∼ 1 x a ∗ ∗   1 ∗ ∗ ∗ ∗ 1 ∗ ∗ ∗ ∗ Indeed, the above two equations lead to 9 possible cases, the first of which is, as desired, a = b and x = y. As for the remaining 8 cases, here once again things are determined by 2 parameters, and in practice, we can always permute the first 3 rows and 3 columns, and then dephase our matrix, as for our matrix to take the above special form. With this result in hand, the combinatorics of the scalar products between the first 3 rows, and between the first 3 columns as well, becomes something which is quite simple HADAMARD MATRICES 31 to investigate. By doing a routine study here, and then completing it with a study of the lower right 2 × 2 corner as well, we are led to 2 possible cases, as follows: 1 1 1 1 1 1 1 1 1 1 1 a b c d 1 a b c d     H ∼ 1 b a d c : H ∼ 1 b a d c     1 c d a b 1 c d b a 1 d c b a 1 d c a b Our claim now is that the first case is in fact not possible. Indeed, we must have: a + b + c + d = −1 2Re(a¯b) + 2Re(cd¯) = −1 2Re(ac¯) + 2Re(bd¯) = −1 2Re(ad¯) + 2Re(bc¯) = −1 Since we have |Re(x)| ≤ 1 for any x ∈ T, we deduce from the second equation that Re(a¯b) ≤ 1/2, and so that the arc length between a, b satisfies θ(a, b) ≥ π/3. The same argument applies to c, d, and to the other pairs of numbers in the last 2 equations. Now since our equations are invariant under permutations of a, b, c, d, we can assume that a, b, c, d are ordered on the circle, and by the above, separated by ≥ π/3 arc lengths. But this implies θ(a, c) ≥ 2π/3 and θ(b, d) ≥ 2π/3, which gives Re(ac¯) ≤ −1/2 and Re(bd¯) ≤ −1/2, which contradicts the third equation. Thus, our claim is proved. Summarizing, we have proved so far that our matrix must be as follows: 1 1 1 1 1 1 a b c d   H ∼ 1 b a d c   1 c d b a 1 d c a b We are now in position of finishing. The orthogonality equations are as follows: a + b + c + d = −1 2Re(a¯b) + 2Re(cd¯) = −1 ac¯ + c¯b + bd¯+ da¯ = −1 The third equation can be written in the following equivalent form: Re[(a + b)(¯c + d¯)] = −1 Im[(a − b)(¯c − d¯)] = 0 a+b c+d From a, b, c, d ∈ T we obtain a−b , c−d ∈ iR, so we can find s, t ∈ R such that: a + b = is(a − b) , c + d = it(c − d) 32 TEO BANICA

By plugging in these values, our system of equations simpliﬁes, as follows: (a + b) + (c + d) = −1 |a + b|2 + |c + d|2 = 3 (a + b)(¯c + d¯) = −1 Now observe that the last equation implies in particular that we have: |a + b|2 · |c + d|2 = 1 Thus |a + b|2, |c + d|2 must be roots of X2 − 3X + 1 = 0, which gives: (√ √ ) n o 5 + 1 5 − 1 |a + b| , |c + d| = , 2 2 This is very good news, because we are now into 5-th roots of unity. To be more precise, we have 2 cases to be considered, the ﬁrst one being as follows, with z ∈ T: √ √ 5 + 1 5 − 1 a + b = z , c + d = − z 2 2 From a + b + c + d = −1 we obtain z = −1, and by using this we obtain b =a, ¯ d =c ¯, and then Re(a) = cos(2π/5), Re(c) = cos(π/5), and so we have H ∼ F5. The second case, with a, b and c, d interchanged, this leads to H ∼ F5 as well. The above result is of course something quite impressive. However, at the level of practical conclusions, we can only say that the N = 5 case is something very simple. At N = 6 now, the situation becomes considerably complicated, with lots of “exotic” solutions, and with the structure of the Hadamard manifold X6 being not understood yet. In fact, this manifold X6 looks as complicated as real algebraic manifolds can get. The simplest examples of Hadamard matrices at N = 6 are as follows: Theorem 2.14. We have the following basic Hadamard matrices, at N = 6:

(1) The Fourier matrix F6. (2) The Dit¸˘adeformations of F2 ⊗ F3 and of F3 ⊗ F2. q (3) The Haagerup matrix H6 . (4) The Tao matrix T6. Proof. All this is elementary, the idea, and formulae of the matrices, being as follows: (1) This is something that we know well. (2) Consider indeed the dephased Dit¸˘adeformations of F2 ⊗ F3 and F3 ⊗ F2: (rs) (r) F = F ⊗  F ,F s = F ⊗ F 6 2 1 1 1 3 6 3 1 1 2     1 r s 1 r   1 s HADAMARD MATRICES 33

Here r, s are two parameters on the unit circle, r, s ∈ T. In matrix form: 1 1 1 1 1 1  2 2 1 w w 1 w w   2 2  1 w w 1 w w  (rs)   F6 =   1 r s −1 −r −s    1 wr w2s −1 −wr −w2s 1 w2r ws −1 −w2r −ws As for the other deformation, this is given by: 1 1 1 1 1 1  1 −1 1 −1 1 −1     2 2  r 1 r w wr w w r  (s)   F6 =  2 2  1 −r w −wr w −w r     1 s w2 w2s w ws  1 −s w2 −w2s w −ws

(3) The matrix here, from [51], is as follows, with q ∈ T: 1 1 1 1 1 1  1 −1 i i −i −i   q 1 i −1 −i q −q H6 =   1 i −i −1 −q q  1 −i q¯ −q¯ i −1 1 −i −q¯ q¯ −1 i (4) The matrix here, from [90], is as follows, with w = e2πi/3: 1 1 1 1 1 1  2 2 1 1 w w w w   2 2  1 w 1 w w w  T6 =  2 2 1 w w 1 w w  1 w2 w2 w 1 w  1 w2 w w2 w 1 q Observe that both H6 and T6 are indeed complex Hadamard matrices. The point with the matrices in Theorem 2.14 is that they are “regular”, in the sense that the scalar products between rows appear in the simplest possible way, namely from vanishing sums of roots of unity, possibly rotated by a scalar. We will be back to this in section 3 below, with a proof that these matrices are the only regular ones, at N = 6. 34 TEO BANICA

In the non-regular case now, there are many known constructions at N = 6. Here is one such construction, mildly “exotic”, found by Björck and Fröberg in [31]: Proposition 2.15. The following is a complex Hadamard matrix,  1 ia −a −i −a¯ ia¯   ia¯ 1 ia −a −i −a¯   −a¯ ia¯ 1 ia −a −i BF6 =   −i −a¯ ia¯ 1 ia −a −a −i −a¯ ia¯ 1 ia  ia −a −i −a¯ ia¯ 1 √ where a ∈ T is one of the roots of a2 + ( 3 − 1)a + 1 = 0. Proof. Observe that the matrix in the statement is circulant, in the sense the rows appear by cyclically permuting the first row. Thus, we only have to√ check that the first row is 2 orthogonal to the other 5 rows. But this follows from a + ( 3 − 1)a + 1 = 0. The obvious question here is perhaps on how Björck and Fröberg were able to construct the above matrix (!) This was done via some general theory for the circulant Hadamard matrices, and some computer simulations. We will discuss this in section 6 below. Further study in the N = 6 case leads to a number of horrors, of real algebraic geometric flavor, and we have here, as an illustrating example, the following result from [25]: Theorem 2.16. The self-adjoint 6 × 6 Hadamard matrices are, up to equivalence 1 1 1 1 1 1  1 −1x ¯ −y −x¯ y    q 1 x −1 t −t −x BN6 =  ¯ ¯ 1 −y¯ t −1y ¯ −t 1 −x −t¯ y 1z ¯  1y ¯ −x¯ −t z 1 with x, y, z, t ∈ T depending on a parameter q ∈ T, in a very complicated way. Proof. The study here can be done via a lot of work, and tricks, in the spirit of the Haagerup classification result at N = 5, and the equations are as follows: √ 1 + 2q + q2 − 2p1 + 2q + 2q3 + q4 x = 1 + 2q − q2 y = q 1 + 2q − q2 z = q(−1 + 2q + q2) √ 1 + 2q + q2 − 2p1 + 2q + 2q3 + q4 t = −1 + 2q + q2 All this is very technical, and we refer here to [25]. HADAMARD MATRICES 35

There are many other examples at N = 6, and no classiﬁcation known. See [68]. Let us discuss now the case N = 7. We will restrict the attention to case where the combinatorics comes from roots of unity. We use the following result, from [87]:

Theorem 2.17. If H ∈ MN (±1) with N ≥ 8 is dephased symmetric Hadamard, and √ (1 ± i N − 5)2 w = N − 4 then the following procedure yields a complex Hadamard matrix M ∈ MN−1(T): (1) Erase the first row and column of H. (2) Replace all diagonal 1 entries with −w. (3) Replace all off-diagonal −1 entries with w. Proof. We know from the proof of Proposition 1.8 that the scalar product between any two rows of H, normalized as there, appears as follows: N N N N P = · 1 · 1 + · 1 · (−1) + · (−1) · 1 + · (−1) · (−1) = 0 4 4 4 4 Let us peform now the above operations (1,2,3), in reverse order. When replacing −1 → w, all across the matrix, the above scalar product becomes: N N N N N P 0 = · 1 · 1 + · 1 · w¯ + · w · 1 + · (−1) · (−1) = (1 + Re(w)) 4 4 4 4 2 Now when adjusting the diagonal via w → −1 back, and 1 → −w, this amounts in adding the quantity −2(1 + Re(w)) to our product. Thus, our product becomes: N N − 4 6 − N P 00 = − 2 (1 + Re(w)) = 1 + = 1 2 2 N − 4 Finally, erasing the first row and column amounts in substracting 1 from our scalar 000 product. Thus, our scalar product becomes P = 1 − 1 = 0, and we are done. Observe that the number w in the above statement is a root of unity precisely at N = 8, where the only matrix satisfying the conditions in the statement is the Walsh matrix W8. So, let us apply, as in [87], the above construction to this matrix: 1 1 1 1 1 1 1 1  ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗  1 −1 1 −1 1 −1 1 −1 ∗ −1 1 w 1 w 1 w      1 1 −1 −1 1 1 −1 −1 ∗ 1 −1 w 1 1 w w      1 −1 −1 1 1 −1 −1 1  ∗ w w −w 1 w w 1    →   1 1 1 1 −1 −1 −1 −1 ∗ 1 1 1 −1 w w w      1 −1 1 −1 −1 1 −1 1  ∗ w 1 w w −w w 1      1 −1 −1 −1 −1 −1 1 1  ∗ 1 w w w w −w 1  1 −1 −1 1 −1 1 1 −1 ∗ w w 1 w 1 1 −1

The matrix on the right is the Petrescu matrix P7, found in [77]. Thus, we have: 36 TEO BANICA

Theorem 2.18. P7 is the unique matrix formed by roots of unity that can be obtained by the Sz¨oll˝osiconstruction. It appears at N = 8, from H = W8. Its formula is  −w if (ijk) = (abc), ia + jb + kc = 0(2)  (P7)ijk,abc = w if (ijk) 6= (abc), ia + jb + kc 6= 0(2) (−1)ia+jb+kc otherwise where w = e2πi/3, and with the indices belonging to the set {0, 1}3 − {(0, 0, 0)}. ij Proof. We know that the Sz¨oll˝osiconstruction maps W8 → P7. Since (F2)ij = (−1) , we ia+jb+kc have (W8)ijk,abc = (−1) , and this gives the formula in the statement.

Now observe that we are in the quite special situation H = F2 ⊗ K, with K being dephased and symmetric. Thus, we can search for a one-parameter aﬃne deformation K(q) which is dephased and symmetric, and then build the following matrix: K(q) K H(q) = K −K(¯q)

In our case, such a deformation K(q) = W4(q) can be obtained by putting the q parameters in the 2 × 2 middle block. Now by performing the Sz¨oll˝osiconstruction, with the parameters q, q¯ left untouched, we obtain the parametric Petrescu matrix [77]: Theorem 2.19. The following is a complex Hadamard matrix, −q q w 1 w 1 w   q −q w 1 1 w w     w w −w 1 w w 1  q  1 1 1 −1 w w w  P7 =    w 1 w w −qw¯ qw¯ 1     1 w w w qw¯ −qw¯ 1  w w 1 w 1 1 −1 where w = e2πi/3, and q ∈ T. Proof. This follows from the above considerations, or from a direct veriﬁcation of the 2 orthogonality of the rows, which uses either 1 − 1 = 0, or 1 + w + w = 0. q Observe that the above matrix P7 has the property of being “regular”, in the sense that the scalar products between rows appear from vanishing sums of roots of unity, possibly rotated by a scalar. We will be back to this in the next section, with the conjectural q statement that F7,P7 are the only regular Hadamard matrices at N = 7. HADAMARD MATRICES 37

3. Roots of unity

Many interesting examples of complex Hadamard matrices H ∈ MN (T), including the real ones H ∈ MN (±1), have as entries roots of unity, of finite order. We discuss here this case, and more generally the “regular” case, where the combinatorics of the scalar products between the rows comes from vanishing sums of roots of unity. Let us begin with the following definition, going back to the work in [34]: Definition 3.1. An Hadamard matrix is called of Butson type if its entries are roots of unity of finite order. The Butson class HN (l) consists of the Hadamard matrices

H ∈ MN (Zl) where Zl is the group of the l-th roots of unity. The level of a Butson matrix H ∈ MN (T) is the smallest integer l ∈ N such that H ∈ HN (l). As basic examples, we have the real Hadamard matrices, which form by deﬁnition the Butson class HN (2). The Fourier matrices are Butson matrices as well, because we have FN ∈ HN (N), and more generally FG ∈ HN (l), with N = |G|, and with l ∈ N being the smallest common order of the elements of G. There are many other examples of such matrices, as for instance those in Theorem 2.14, at 1 values of the parameters. Generally speaking, the main question regarding the Butson matrices is that of understanding when HN (l) 6= 0, via a theorem providing obstructions, and then a conjecture stating that these obstructions are the only ones. Let us begin with: Proposition 3.2 (Sylvester obstruction). The following holds,

HN (2) 6= ∅ =⇒ N ∈ {2} ∪ 4N due to the orthogonality of the ﬁrst 3 rows. Proof. This is something that we know from section 1, with the obstruction, going back to Sylvester’s paper [85], being explained in Proposition 1.8 above. The above obstruction is fully satisfactory, because according to the Hadamard Con- jecture, its converse should hold. Thus, we are fully done with the case l = 2. Our purpose now will be that of ﬁnding analogous statements at l ≥ 3, theorem plus conjecture. At very small values of l this is certainly possible, and in what regards the needed obstructions, we can get away with the following simple fact, from [34], [98]: Proposition 3.3. For a prime power l = pa, the vanishing sums of l-th roots of unity

λ1 + ... + λN = 0 , λi ∈ Zl appear as formal sums of rotated full sums of p-th roots of unity. 38 TEO BANICA

Proof. Consider indeed the full sum of p-th roots of unity, taken in a formal sense: p X S = (e2πi/p)k k=1 2πi/l r r Let also w = e , and for r ∈ {1, 2, . . . , l/p} denote by Sp = w · S the above sum, rotated by wr. We must show that any vanishing sum of l-th roots of unity appears as a r sum of such quantities Sp, with all this taken of course in a formal sense. For this purpose, consider the following map, which assigns to the abstract elements of the group ring Z[Zl] their precise numeric values, inside Z(w) ⊂ C:

Φ: Z[Zl] → Z(w) r Our claim is that the elements {Sp} form a basis of ker Φ. Indeed, we obviously have r r Sp ∈ ker Φ. Also, these elements are linearly independent, because the support of Sp a−1 contains a unique element of the subset {1, 2, . . . , p } ⊂ Zl, namely the element r ∈ Zl, r so all the coefficients of a vanishing linear combination of sums Sp must vanish. r Thus, we are left with proving that ker Φ is spanned by {Sp}. For this purpose, let us recall that the minimal polynomial of w is as follows: pa X − 1 a−1 a−1 a−1 = 1 + Xp + X2p + ... + X(p−1)p Xpa−1 − 1 But this shows that ker Φ has dimension pa − (pa − pa−1) = pa−1, and since this is r exactly the number of the sums Sp, this finishes the proof of our claim. P r Thus, any vanishing sum of l-th roots of unity must be of the form ±Sp, and the above support considerations show the coefficients must be positive, as desired. We can now formulate a result in the spirit of Proposition 3.2, as follows: Proposition 3.4 (Butson obstruction). The following holds, a HN (p ) 6= ∅ =⇒ N ∈ pN due to the orthogonality of the first 2 rows. Proof. This follows indeed from Proposition 3.3, because the scalar product between the first 2 rows of our matrix is a vanishing sum of l-th roots of unity. WIth these obstructions in hand, we can discuss the case l ≤ 5, as follows: Theorem 3.5. We have the following results,

(1) HN (2) 6= ∅ =⇒ N ∈ {2} ∪ 4N, (2) HN (3) 6= ∅ =⇒ N ∈ 3N, (3) HN (4) 6= ∅ =⇒ N ∈ 2N, (4) HN (5) 6= ∅ =⇒ N ∈ 5N, with in cases (1,3), a solid conjecture stating that the converse should hold as well. HADAMARD MATRICES 39

Proof. In this statement (1) is the Sylvester obstruction, and (2,3,4) are particular cases of the Butson obstruction. As for the last assertion, which is of course something rather informal, but which is important for our purposes, the situation is as follows: (1) Here, as already mentioned, we have the Hadamard Conjecture, which comes with very solid evidence, as explained in section 1 above. (2) Here we have an old conjecture, dealing with complex Hadamard matrices over {±1, ±i}, going back to the work in [93], and called Turyn Conjecture. At l = 3 the situation is quite complicated, due to the following result, from [40]: Proposition 3.6 (de Launey obstruction). The following holds,

2πi/l 2 N HN (l) 6= ∅ =⇒ ∃ d ∈ Z[e ], |d| = N due to the orthogonality of all N rows. In particular, we have

5|N =⇒ HN (6) = ∅ and so H15(3) = ∅, which shows that the Butson obstruction is too weak at l = 3. Proof. The obstruction follows from the unitarity condition HH∗ = N for the complex Hadamard matrices, by applying the determinant to it, which gives: |det(H)|2 = N N Regarding the second assertion, let w = e2πi/3, and assume that d = a + bw + cw2 with a, b, c ∈ Z satisﬁes |d|2 = 0(5). We have the following computation: |d|2 = (a + bw + cw2)(a + bw2 + cw) = a2 + b2 + c2 − ab − bc − ac 1 = [(a − b)2 + (b − c)2 + (c − a)2] 2 Thus our condition |d|2 = 0(5) leads to the following system, modulo 5: x + y + z = 0 , x2 + y2 + z2 = 0 But this system has no solutions. Indeed, let us look at x2 +y2 +z2 = 0. If this equality appears as 0 + 0 + 0 = 0 we can divide x, y, z by 5 and redo the computation, and if not, this equality can only appear as 0 + 1 + (−1) = 0. Thus, modulo permutations, we must have x = 0, y = ±1, z = ±2, which contradicts x + y + z = 0. Finally, the last assertion follows from H15(3) ⊂ H15(6) = ∅. At l = 5 now, things are a bit unclear, with the converse of Theorem 3.5 (4) being something viable, at the conjectural level, at least to our knowledge. At l = 6 the situation becomes again complicated, as follows: 40 TEO BANICA

Proposition 3.7 (Haagerup obstruction). The following holds, due to Haagerup’s N = 5 classiﬁcation result, involving the orthogonality of all 5 rows of the matrix:

H5(l) 6= ∅ =⇒ 5|l

In particular we have H5(6) = ∅, which follows by the way from the de Launey obstruction as well, in contrast with the fact that we generally have HN (6) 6= ∅.

Proof. In this statement the obstruction H5(l) = ∅ =⇒ 5|l comes indeed from Haagerup’s classiﬁcation result, explained in Theorem 2.13 above. As for the last assertion, this is something very informal, the situation at small values of N being as follows: – At N = 2, 3, 4 we have the matrices F2,F3,W4. 1 – At N = 6, 7, 8, 9 we have the matrices F6,P7 ,W8,F3 ⊗ F3. – At N = 10 we have the following matrix, found in [13] by using a computer, and written in logarithmic form, with k standing for e2kπi/6:  0 0 0 0 0 0 0 0 0 0   0 4 1 5 3 1 3 3 5 1     0 1 2 3 5 5 1 3 5 3     0 5 3 2 1 5 3 5 3 1    6  0 3 5 1 4 1 1 5 3 3  X10 =    0 3 3 3 3 3 0 0 0 0   0 1 1 5 3 4 3 0 2 4     0 1 5 3 5 2 4 3 2 0     0 5 3 5 1 2 0 2 3 4  0 3 5 1 1 4 4 2 0 3

We refer to [13] for more details on this topic.

All this is not good news. Indeed, there is no hope of conjecturally solving our HN (l) 6= ∅ problem in general, because this would have to take into account, and in a simple and conceptual way, both the subtle arithmetic consequences of the de Launey obstruction, and the Haagerup classiﬁcation result at N = 5, and this is something not feasible.

In order to further comment on these diﬃculties, let us discuss now a generalization of Proposition 3.3 above, and of the related Butson obstruction from Proposition 3.4, which has been our main source of obstructions, so far. Let us start with: Deﬁnition 3.8. A cycle is a full sum of roots of unity, possibly rotated by a scalar,

l X k 2πi/l C = q w , w = e , q ∈ T k=1 and taken in a formal sense. A sum of cycles is a formal sum of cycles. HADAMARD MATRICES 41

The actual sum of a cycle, or of a sum of cycles, is of course 0. This is why the word “formal” is there, for reminding us that we are working with formal sums. As an example, here is a sum of cycles, with w = e2πi/6, and with |q| = 1: 1 + w2 + w4 + qw + qw4 = 0 We know from Proposition 3.3 above that any vanishing sum of l-th roots of unity must be a sum of cycles, at least when l = pa is a prime power. However, this is not the case in general, the simplest counterexample being as follows, with w = e2πi/30: w5 + w6 + w12 + w18 + w24 + w25 = 0 The following deep result on the subject is due to Lam and Leung [66]:

a1 ak Theorem 3.9. Let l = p1 . . . pk , and assume that λi ∈ Zl satisfy λ1 + ... + λN = 0. P (1) λi is a sum of cycles, with Z coefficients. P (2) If k ≤ 2 then λi is a sum of cycles (with N coefficients). P (3) If k ≥ 3 then λi might not decompose as a sum of cycles. P (4) λi has the same length as a sum of cycles: N ∈ p1N + ... + pkN. Proof. This is something that we will not really need in what follows, but that we included here, in view of its importance. The idea of the proof is as follows: (1) This is a well-known result, which follows from basic number theory, by using arguments in the spirit of those in the proof of Proposition 3.3 above. (2) This is something that we already know at k = 1, from Proposition 3.3. At k = 2 the proof is more technical, along the same lines. See [66]. (3) The smallest possible l potentially producing a counterexample is l = 2 · 3 · 5 = 30, and we have here indeed the sum given above, with w = e2πi/30. (4) This is a deep result, due to Lam and Leung, relying on advanced number theory knowledge. We refer to their paper [66] for the proof. As a consequence of the above result, we have the following generalization of the Butson obstruction, which is something final and optimal on this subject:

a1 ak Theorem 3.10 (Lam-Leung obstruction). Assuming l = p1 . . . pk , the following must hold, due to the orthogonality of the ﬁrst 2 rows:

HN (l) 6= ∅ =⇒ N ∈ p1N + ... + pkN In the case k ≥ 2, the latter condition is automatically satisfied at N >> 0. Proof. Here the first assertion, which generalizes the l = pa obstruction from Proposition 3.4 above, comes from Theorem 3.9 (4), applied to the vanishing sum of l-th roots of unity coming from the scalar product between the first 2 rows. As for the second assertion, this is something well-known, coming from basic number theory. 42 TEO BANICA

Summarizing, our study so far of the condition HN (l) 6= ∅ has led us into an optimal obstruction coming from the first 2 rows, namely the Lam-Leung one, then an obstruction coming from the first 3 rows, namely the Sylvester one, and then two subtle obstructions coming from all N rows, namely the de Launey one, and the Haagerup one. As an overall conclusion, by contemplating all these obstructions, nothing good in relation with our problem HN (l) 6= ∅ is going on at small N. So, as a natural and more modest objective, we should perhaps try instead to solve this problem at N >> 0. The point indeed is that everything simplifies at N >> 0, with some of the above obstructions dissapearing, and with some other known obstructions, not to be discussed here, dissapearing as well. We are therefore led to the following statement: Conjecture 3.11 (Asymptotic Butson Conjecture (ABC)). The following equivalences should hold, in an asymptotic sense, at N >> 0,

(1) HN (2) 6= ∅ ⇐⇒ 4|N, a a (2) HN (p ) 6= ∅ ⇐⇒ p|N, for p ≥ 3 prime power, (3) HN (l) 6= ∅ ⇐⇒ ∅, for l ∈ N not a prime power, modulo the de Launey obstruction, |d|2 = N N for some d ∈ Z[e2πi/l]. In short, our belief is that when imposing the condition N >> 0, only the Sylvester, Butson and de Launey obstructions survive. This is of course something quite nice, but in what regards a possible proof, there is probably no way. Indeed, our above conjecture generalizes the HC in the N >> 0 regime, which is something beyond reach. One interesting idea, however, in dealing with such questions, coming from the de Launey-Levin result from [44], explained in section 1, is that of looking at the partial Butson matrices, at N >> 0. Observe in particular that restricting the attention to the rectangular case, and this not even in the N >> 0 regime, would make dissapear the de Launey obstruction from the ABC, which uses the orthogonality of all N rows. We will discuss this later on, at the end of this section. For a number of related considerations, we refer as well to the papers [40], [43]. Getting away now from all this arithmetic madness, let us discuss now, as a more concrete thing, the classiﬁcation of the regular complex Hadamard matrices of small order. The deﬁnition here, which already appeared in the above, is as follows:

Deﬁnition 3.12. A complex Hadamard matrix H ∈ MN (T) is called regular if the scalar products between rows decompose as sums of cycles. We should mention that there is some notational clash here, with this notion being sometimes used in order to designate the bistochastic matrices. In this book we use the above notion of regularity, and we call bistochastic the bistochastic matrices. Our purpose in what follows will be that of showing that the notion of regularity can lead to full classiﬁcation results at N ≤ 6, and perhaps at N = 7 too, and all this while covering most of the interesting complex Hadamard matrices that we met, so far. HADAMARD MATRICES 43

As a ﬁrst observation, supporting this last claim, we have the following result: Proposition 3.13. The following complex Hadamard matrices are regular: s (1) The matrices at N ≤ 5, namely F2,F3,F4 ,F5. r (rs) (s) q (2) The main examples at N = 6, namely F6 ,F6 ,H6 ,T6. q (3) The main examples at N = 7, namely F7,P7 .

Proof. The Fourier matrices FN are all regular, with the scalar products between rows appearing as certain sums of full sums of l-th roots of unity, with l|N. As for the other matrices appearing in the statement, with the convention that “cycle structure” means the length of the cycles in the regularity property, the situation is as follows: s (1) F4 has cycle structure 2 + 2, and this because the verification of the Hadamard condition is always based on the formula 1 + (−1) = 0, rotated by scalars. r (rs) (s) (2) F6 ,F6 have mixed cycle structure 2 + 2 + 2/3 + 3, in the sense that both cases q appear, H6 has cycle structure 2 + 2 + 2, and T6 has cycle structure 3 + 3. q 2 (3) P7 has cycle structure 3+2+2, its Hadamard property coming from 1+w+w = 0, 2πi/3 with w = e , and from 1 + (−1) = 0, applied twice, rotated by scalars. Let us discuss now the classification of regular matrices. We first have: Theorem 3.14. The regular Hadamard matrices at N ≤ 5 are s F2,F3,F4 ,F5 up to the equivalence relation for the complex Hadamard matrices. Proof. This is something that we already know, coming from the classification results from section 2, and from Proposition 3.13 (1). However, and here comes our point, proving this result does not need in fact all this, the situation being as follows: (1) At N = 2 the cycle structure can be only 2, and we obtain F2. (2) At N = 3 the cycle structure can be only 3, and we obtain F3. s (3) At N = 4 the cycle structure can be only 2 + 2, and we obtain F4 . (4) At N = 5 some elementary combinatorics shows that the cycle structure 3 + 2 is excluded. Thus we are left with the cycle structure 5, and we obtain F5. Let us discuss now the classification at N = 6. The result here, from [13], states that r (rs) (s) q the above matrices F6 ,F6 ,H6 ,T6 are the only solutions. The proof of this fact is quite long and technical, but we will present here its main ideas. Let us start with: Proposition 3.15. The regular Hadamard matrices at N = 6 fall into 3 classes:

(1) Cycle structure 3 + 3, with T6 being an example. q (2) Cycle structure 2 + 2 + 2, with H6 being an example. r (rs) (s) (3) Mixed cycle structure 3 + 3/2 + 2 + 2, with F6 ,F6 being examples. Proof. This is a bit of an empty statement, with the above (1,2,3) possibilities being the only ones, and with the various examples coming from Proposition 3.13 (2). 44 TEO BANICA

In order to do the classiﬁcation, we must prove that the examples in (1,2,3) are the only ones. Let us start with the Tao matrix. The result here is as follows:

Proposition 3.16. The matrix T6 is the only one with cycle structure 3 + 3. Proof. The proof of this fact, from [13], is quite long and technical, the idea being that of studying first the 3 × 6 case, then the 4 × 6 case, and finally the 6 × 6 case. So, consider first a partial Hadamard matrix A ∈ M3×6(T), with the scalar products between rows assumed to be all of type 3+3. By doing some elementary combinatorics, one 2 can show that, modulo equivalence, either all the entries of A belong to Z3 = {1, w, w }, or A has the following special form, for certain parameters r, s ∈ T: 1 1 1 1 1 1  A = 1 w w2 r wr w2r 1 w2 w s w2s ws With this in hand, we can now investigate the 4×6 case. Assume indeed that we have a partial Hadamard matrix B ∈ M4×6(T), with the scalar products between rows assumed to be all of type 3 + 3. By looking at the 4 submatrices A(1),A(2),A(3),A(4) obtained from B by deleting one row, and applying the above 3 × 6 result, we are led, after doing some combinatorics, to the conclusion that all the possible parameters dissapear:

B ∈ M4×6(Z3) With this result in hand, we can go now for the general case. Indeed, an Hadamard matrix M ∈ M6(T) having cycle structure 3 + 3 must be as follows:

M ∈ M6(Z3)

But the study here is elementary, with T6 as the only solution. See [13]. Regarding now the Haagerup matrix, the result is similar, as follows:

q Proposition 3.17. The matrix H6 is the only one with cycle structure 2 + 2 + 2. Proof. The proof here, from [13], uses the same idea as in the proof of Proposition 3.16. The study of the 3 × 6 partial Hadamard matrices with cycle structure 2 + 2 + 2 leads, up to equivalence, to the following 4 solutions, with q ∈ T being a parameter: 1 1 1 1 1 1  A1 = 1 −i 1 i −1 −1 1 −1 i −i q −q

1 1 1 1 1 1  A2 = 1 1 −1 i −1 −i  1 −1 q −q iq −iq HADAMARD MATRICES 45

1 1 1 1 1 1  A3 = 1 −1 i −i q −q 1 −i i −1 −q q

1 1 1 1 1 1  A4 = 1 −i −1 i q −q 1 −1 −q −iq iq q With this result in hand, we can go directly for the 6 × 6 case. Indeed, a careful examination of the 3×6 submatrices, and of the way that diﬀerent parameters can overlap vertically, shows that our matrix must have a 3 × 3 block decomposition as follows: ABC  M = D xE yF  G zH tI Here A, . . . , I are 2 × 2 matrices over {±1, ±i}, and x, y, z, t are in {1, q}. A more careful examination shows that the solution must be of the following form: ABC  M = D E qF  G qH qI More precisely, the matrix must be as follows: 1 1 1 1 1 1  1 1 −i i −1 −1    1 i −1 −i −q q  M =   1 −i i −1 −iq iq  1 −1 q −iq iq −q  1 −1 −q iq q −iq

q But this matrix is equivalent to H6 , and we are done. See [13]. Regarding now the mixed case, where both 2 + 2 + 2 and 3 + 3 situations can appear, this is a bit more complicated, and requires some preliminary discussion. We can associate to any mixed Hadamard matrix M ∈ M6(C) its “row graph”, having the 6 rows as vertices, and with each edge being called “binary” or “ternary”, depending on whether the corresponding scalar product is of type 2 + 2 + 2 or 3 + 3. With this convention, we have the following result:

Proposition 3.18. The row graph of a mixed matrix M ∈ M6(C) can be: (1) Either the bipartite graph having 3 binary edges. (2) Or the bipartite graph having 2 ternary triangles. 46 TEO BANICA

Proof. This is once again something a bit technical, from [13], the idea being as follows. Let X be the row graph in the statement. By doing some combinatorics, quite long but of very elementary type, we are led to the following conclusions about X: – X has no binary triangle. – X has no ternary square. – X has at least one ternary triangle. With these results in hand, we see that there are only two types of squares in our graph X, namely those having 1 binary edge and 5 ternary edges, and those consisting of a ternary triangle, connected to the 4-th point with 3 binary edges. By looking at pentagons, then hexagons that can be built with these squares, we see that the above two types of squares cannot appear at the same time, at that at the level of hexagons, we have the two solutions in the statement. See [13]. We can now complete our classiﬁcation at N = 6, as follows:

r (rs) (s) Proposition 3.19. The matrices F6 ,F6 are the only ones with mixed cycle structure. Proof. According to Proposition 3.18, we have two cases: (1) Assume first that the row graph is the bipartite one with 3 binary edges. By permuting the rows, the upper 4 × 6 submatrix of our matrix must be as follows: 1 1 1 1 1 1  1 w w2 r wr w2r B =  2 2  1 w w s w s ws  1 1 1 t t t Now since the scalar product between the first and the fourth row is binary, we must have t = −1, so the solution is: 1 1 1 1 1 1  1 w w2 r wr w2r B =  2 2  1 w w s w s ws  1 1 1 −1 −1 −1 We can use the same argument for finding the fifth and sixth row, by arranging the matrix formed by the first three rows such as the second, respectively third row consist only of 1’s. This will make appear some parameters of the form w, w2, r, s in the extra (rs) row, and we obtain in this way a matrix which is equivalent to F6 . See [13]. (2) Assume now that the row graph is the bipartite one with 2 ternary triangles. By permuting the rows, the upper 4 × 6 submatrix of our matrix must be as follows: 1 1 1 1 1 1  1 1 w w w2 w2  B =  2 2  1 1 w w w w  1 −1 r −r s −s HADAMARD MATRICES 47

We can use the same argument for ﬁnding the ﬁfth and sixth row, and we conclude that the matrix is of the following type: 1 1 1 1 1 1  2 2 1 1 w w w w   2 2  1 1 w w w w  M =   1 −1 r −r s −s 1 −1 a −a b −b 1 −1 c −c d −d Now since the last three rows must form a ternary triangle, we conclude that the matrix must be of the following form: 1 1 1 1 1 1  2 2 1 1 w w w w   2 2  1 1 w w w w  M =   1 −1 r −r s −s  1 −1 wr −wr w2s −w2s 1 −1 w2r −w2r ws −ws

r (s) But this matrix is equivalent to F6 , and we are done. See [13]. Summing up all the above, we have proved the following theorem: Theorem 3.20. The regular complex Hadamard matrices at N = 6 are: r (rs) (s) (1) The deformations F6 ,F6 of the Fourier matrix F6. q (2) The Haagerup matrix H6 . (3) The Tao matrix T6. Proof. This follows indeed from the trichotomy from Proposition 3.15, and from the results in Proposition 3.16, Proposition 3.17 and Proposition 3.19. See [13]. All this is quite nice, and our belief is that the N = 7 classiﬁcation is doable as well. Here we have 3 possible cycle structures, namely 3 + 2 + 2, 5 + 2, 7, and some elementary number theory shows that 5 + 2 is excluded, and that 3 + 2 + 2 and 7 cannot interact. Thus we have a dichotomy, and our conjecture is as follows: Conjecture 3.21. The regular complex Hadamard matrices at N = 7 are:

(1) The Fourier matrix F7. q (2) The Petrescu matrix P7 .

Regarding (1), one can show indeed that F7 is the only matrix having cycle structure 7, with this being related to some more general results from [55]. As for (2), the problem q is that of proving that P7 is the only matrix having cycle structure 3 + 2 + 2. The computations here are unfortunately far more involved than those at N = 6, briefly presented above, and finishing the classification work here is not an easy question. 48 TEO BANICA

As a conclusion to all this, when imposing the regularity condition, things simplify a bit, with respect to the general case, according to a kind of N → N + 1 rule. To be more precise, the diﬃculties in the general case are basically of real algebraic geometry nature, and can be labeled as easy at N ≤ 4, hard at N = 5, and not solved yet at N = 6. As for the regular case, here the diﬃculties are basically of design theory nature, and can be labeled as easy at N ≤ 5, hard at N = 6, and not solved yet at N = 7.

Besides the classiﬁcation questions, there are as well a number of theoretical questions in relation with the notion of regularity, that we believe to be very interesting. We have for instance the following conjecture, going back to [13], and then to [22]: Conjecture 3.22 (Regularity Conjecture). The following hold:

(1) Any Butson matrix H ∈ MN (C) is regular. (2) Any regular matrix H ∈ MN (C) is an affine deformation of a Butson matrix. In other words, the first conjecture is that a “tricky vanishing sum” of roots of unity, like the l = 30 one given after Definition 3.8 above, cannot be used in order to construct a complex Hadamard matrix. This is a quite difficult question, coming however with substantial computer evidence. We have no idea on how to approach it. See [13]. As for the second conjecture, this simply comes from the known examples of regular Hadamard matrices, which all appear from certain Butson matrices, by inserting parameters, in an affine way. This conjecture is from [22], and we will further discuss the notion of affine deformation, with some general results on the subject, in section 4 below.

We would like to end this section, which was depressingly algebraic and diﬃcult, with no simple and conceptual result in sight, by doing some analysis. As explained after Conjecture 3.11 above (ABC), one way of getting into analysis, in connection with root of unity questions, is that of looking at the partial Butson matrices, at N >> 0. The idea here comes of course from the de Launey-Levin counting result from [44], explained in section 1 above. Let us ﬁrst discuss the prime power case. We have: Proposition 3.23. When q = pk is a prime power, the standard form of the dephased partial Butson matrices at M = 2 is  1 1 ... 1 ...... 1 1 ... 1  q/p−1 q−q/p q−q/p+1 q−1 H =  1 w . . . w ...... w w . . . w  |{z} |{z} | {z } | {z } | {z } |{z} a1 a2 aq/p a1 a2 aq/p

2πi/q where w = e and where a1, . . . , aq/p ∈ N are multiplicities, summing up to N/p. k Proof. Indeed, it is well-known that for q = p the solutions of λ1 + ... + λN = 0 with λi ∈ Zq are, up to permutations of the terms, exactly those in the statement. Our objective will be to count the matrices in Proposition 3.23. We will need: HADAMARD MATRICES 49

Proposition 3.24. We have the estimate s p s(p−1) X n pn s ' s s−1 (s−1)(p−1) a1, . . . , as p (2πn) a1+...+as=n in the n → ∞ limit. Proof. This is proved by Richmond and Shallit in [81] at p = 2, and the proof in the general case, p ∈ N, is similar. More precisely, let us denote by csp the sum on the left. n √ By setting ai = s + xi n and then by using the various formulae in [81], we obtain:

csp s ! pn (1−s)p sp sp X 2 ' s (2πn) 2 s 2 exp − x 2 i i=1 Z n Z n s ! (1−s)p sp sp X pn 2 2 2 ' s (2πn) s ... exp − xi da1 . . . das−1 0 0 2 | {z } i=1 s−1  2 Z n Z n s−1 s−1 ! (1−s)p sp s−1 sp X sp X pn 2 2 2 2 = s (2πn) s n ... exp − xi − xi  dx1 . . . dxs−1 0 0 2 2 | {z } i=1 i=1 s−1 1−s pn (1−s)p sp s−1 s−1 − 1 sp 2 = s (2πn) 2 s 2 n 2 × π 2 s 2 2 1−s pn (1−s)p sp − 1 + 1−s p 2 = s (2πn) 2 s 2 2 2 2πn pn (1−s)(p−1) sp−s 1−s = s (2πn) 2 s 2 p 2 Thus we have obtained the formula in the statement, and we are done. Now with Proposition 3.24 in hand, we can prove: Theorem 3.25. When q = pk is a prime power, the probability for a randomly chosen M ∈ M2×N (Zq), with N ∈ pN, N → ∞, to be partial Butson is: s 2− q q− q p p q p P ' 2 q− q (2πN) p

r q/2 q pp k q/2 In particular, for q = p prime, P2 ' (2πN)p−1 . Also, for q = 2 , P2 ' 2 2πN .

Proof. First, the probability PM for a random M ∈ MM×N (Zq) to be PBM is: 1 P = #PBM M qMN M×N 50 TEO BANICA

Thus, according to Proposition 3.23, we have the following formula: 1 X N P2 = N q a1 . . . a1 ...... aq/p . . . aq/p a1+...+aq/p=N/p | {z } | {z } p p p 1 N X N/p = N q N/p . . . N/p a1 . . . aq/p | {z } a1+...+aq/p=N/p p p 1 N 1 X N/p = N × N p N/p . . . N/p (q/p) a1 . . . aq/p | {z } a1+...+aq/p=N/p p Now by using the Stirling formula for the left term, and Proposition 3.24 with s = q/p and n = N/p for the right term, we obtain:

s q (p−1) r pp (q/p) p P = × 2 p−1 q −1 ( q −1)(p−1) (2πN) p p (2πN/p) p s p− q (p−1)− q +1+( q −1)(p−1) q (p−1) p p p p q p = p−1+( q −1)(p−1) (2πN) p s 2− q q− q p p q p = q− q (2πN) p Thus we have obtained the formula in the statement, and we are done. k1 k2 Let us discuss now the case where M = 2 and q = p1 p2 has two prime factors. We ﬁrst examine the simplest such case, namely q = p1p2, with p1, p2 primes:

Proposition 3.26. When q = p1p2 is a product of distinct primes, the standard form of the dephased partial Butson matrices at M = 2 is  1 1 ... 1 ...... 1 1 ... 1  p −1 q−p q−p +1 q−1 H =  1 w . . . w 2 ...... w 2 w 2 . . . w  |{z} |{z} | {z } | {z } | {z } |{z} A11 A12 A1p2 Ap11 Ap12 Ap1p2 2πi/q where w = e , and A ∈ Mp1×p2 (N) is of the form Aij = Bi + Cj, with Bi,Cj ∈ N.

Proof. We use the fact that for q = p1p2 any vanishing sum of q-roots of unity decomposes as a sum of cycles. Now if we denote by Bi,Cj ∈ N the multiplicities of the various p2- cycles and p1-cycles, then we must have Aij = Bi + Cj, as claimed.

Regarding the matrices of type Aij = Bi +Cj, when taking them over integers, Bi,Cj ∈

Z, these form a vector space of dimension p1 + p2 − 1. Given A ∈ Mp1×p2 (Z), the “test” for deciding if we have Aij = Bi + Cj or not is Aij + Akl = Ail + Ajk. HADAMARD MATRICES 51

The problem comes of course from the assumption Bi,Cj ≥ 0, which is quite a subtle one. In what follows we restrict attention to the case p1 = 2. Here we have:

Theorem 3.27. For q = 2p with p ≥ 3 prime, P2 equals the probability for a random walk on Zp to end up on the diagonal, i.e. at a position of type (t, . . . , t), with t ∈ Z.

Proof. According to Proposition 3.26, we must understand the matrices A ∈ M2×p(N) which decompose as Aij = Bi + Cj, with Bi,Cj ≥ 0. But this is an easy task, because depending on A11 vs. A21 we have 3 types of solutions, as follows: a . . . a a . . . a a + t . . . a + t 1 p , 1 p , 1 p a1 . . . ap a1 + t . . . ap + t a1 . . . ap

Here ai ≥ 0 and t ≥ 1. Now since cases 2,3 contribute in the same way, we obtain: 1 X N P2 = N (2p) a1, a1, . . . , ap, ap 2Σai=N 2 X X N + N (2p) a1, a1 + t, . . . , ap, ap + t t≥1 2Σai+pt=N We can write this formula in a more compact way, as follows: 1 X X N P2 = N (2p) a1, a1 + |t|, . . . , ap, ap + |t| t∈Z 2Σai+p|t|=N 1 Now since the sum on the right, when rescaled by (2p)N , is exactly the probability for p a random walk on Z to end up at (t, . . . , t), this gives the result. According to the above result we have P = P P (t), where P (t) with t ∈ is the 2 t∈Z 2 2 Z probability for a random walk on Zp to end up at (t, . . . , t). Observe that, by using Proposition 3.24 above with s, p, n equal respectively to p, 2,N/2, we obtain: 2 (0) 1 N X N/2 P2 = N (2p) N/2 a1, . . . , ap a1+...+ap=N/2 r 2 r pp ' × πN 2p−1(πN)p−1 r p p = 2 2πN (t) Regarding now the probability P2 of ending up at (t, . . . , t), in principle for small t this can be estimated by using a modiﬁcation of the method in [81]. However, it is not clear on how to compute the full diagonal return probability in Theorem 3.27. 52 TEO BANICA

It is possible to establish a few more results in this direction, and we refer here to [7]. However, the main question remains that of adapting the methods in [66] to the root of unity case. As a preliminary observation here, also from [7], we have:

Theorem 3.28. The probability PM for a random H ∈ MM×N (Zq) to be partial Butson equals the probability for a length N random walk with increments drawn from n o M E = (eie¯j)i

Proof. Indeed, with T (e) = (eie¯j)i δΣξi,0 = D e dλ (2π) [−π,π]D M Here D = 2 . The problem is that this formula works when Σξi is real, as is the case in [44], but not when Σξi is complex, as is the case in Theorem 3.28. HADAMARD MATRICES 53

4. Geometry, defect In this section and in the next two ones we discuss various geometric aspects of the complex Hadamard matrices. Let us recall that the complex Hadamard manifold appears as an intersection of smooth real algebraic manifolds, as follows: √ XN = MN (T) ∩ NUN

This intersection is very far from being smooth. Given a point H ∈ XN , the problem is that of understanding the structure of XN around H, which is often singular. There are several ways of discussing this question, a quite straightforward approach, going back to the work in [62], and then in [72], [88], being via the 1-parameter deformations of the complex Hadamard matrices. In what follows we will use an equivalent approach, of more real algebraic geometry ﬂavor, developed in [5], [6].

We denote by Xp an unspeciﬁed neighborhood of a point in a manifold, p ∈ X. Also, r it r itr for q ∈ T1, meaning that q ∈ T is close to 1, we deﬁne q with r ∈ R by (e ) = e . With these conventions, we have the following result:

Proposition 4.1. For H ∈ XN and A ∈ MN (R), the following are equivalent: q Aij (1) Hij = Hijq is an Hadamard matrix, for any q ∈ T1. P ¯ Aik−Ajk (2) k HikHjkq = 0, for any i 6= j and any q ∈ T1. P ¯ (3) k HikHjkϕ(Aik − Ajk) = 0, for any i 6= j and any ϕ : R → C. P ¯ r (4) r HikHjk = 0 for any i 6= j and r ∈ , where E = {k|Aik − Ajk = r}. k∈Eij R ij Proof. These equivalences are all elementary, and can be proved as follows: (1) ⇐⇒ (2) Indeed, the scalar products between the rows of Hq are:

q q X Aik ¯ Ajk X ¯ Aik−Ajk < Hi ,Hj >= Hikq Hjkq¯ = HikHjkq k k (2) =⇒ (4) This follows from the following formula, and from the fact that the power functions {qr|r ∈ R} over the unit circle T are linearly independent:

X ¯ Aik−Ajk X r X ¯ HikHjkq = q HikHjk r k r∈R k∈Eij (4) =⇒ (3) This follows from the following formula: X ¯ X X ¯ HikHjkϕ(Aik − Ajk) = ϕ(r) HikHjk r k r∈R k∈Eij r (3) =⇒ (2) This simply follows by taking ϕ(r) = q . Observe that in the above statement the condition (4) is purely combinatorial. In order to understand the above deformations, which are “aﬃne” in a certain sense, it is convenient to enlarge the attention to all types of deformations. 54 TEO BANICA

We keep using the neighborhood notation Xp introduced above, and we consider functions of type f : Xp → Yq, which by deﬁnition satisfy f(p) = q. With these conventions, let us introduce the following notions:

Deﬁnition 4.2. Let H ∈ MN (C) be a complex Hadamard matrix. (1) A deformation of H is a smooth function f : T1 → (XN )H . Aij (2) The deformation is called “aﬃne” if fij(q) = Hijq , with A ∈ MN (R). ai+bj N (3) We call “trivial” the deformations of type fij(q) = Hijq , with a, b ∈ R .

it iAij t Here the adjective “affine” comes from fij(e ) = Hije , because the function t → Aijt which produces the exponent is indeed affine. As for the adjective “trivial”, this comes ai+bj from the fact that f(q) = (Hijq )ij is obtained from H by multiplying the rows and columns by certain numbers in T, so it is automatically Hadamard. The basic example of an affine deformation comes from the Dit¸˘adeformations H ⊗Q K, by taking all parameters qij ∈ T to be powers of q ∈ T. As an example, here are the exponent matrices coming from the left and right Dit¸˘adeformations of F2 ⊗ F2: a a b b a b a b c c d d a b a b Al =   Ar =   a a b b c d c d c c d d c d c d In order to investigate the above types of deformations, we will use the corresponding tangent vectors. So, let us recall that the manifold XN is given by: √ XN = MN (T) ∩ NUN This observation leads to the following definition, where in the first part we denote by TpX the tangent space to a point in a smooth manifold, p ∈ X: Definition 4.3. Associated to a point H ∈ X are the following objects: N √ (1) The enveloping tangent space: TeH XN = TH MN (T) ∩ TH NUN . (2) The tangent cone TH XN : the set of tangent vectors to the deformations of H. ◦ (3) The affine tangent cone TH XN : same as above, using affine deformations only. × (4) The trivial tangent cone TH XN : as above, using trivial deformations only. × ◦ Observe that TeH XN ,TH XN are real linear spaces, and that TH XN ,TH XN are two-sided cones, in the sense that they satisfy the following condition: λ ∈ R,A ∈ T =⇒ λA ∈ T Observe also that we have inclusions of cones, as follows: × ◦ TH XN ⊂ TH XN ⊂ TH XN ⊂ TeH XN In more algebraic terms now, these various tangent cones are best described by the corresponding matrices, and we have here the following result: HADAMARD MATRICES 55

× ◦ Theorem 4.4. The cones TH XN ⊂ TH XN ⊂ TH XN ⊂ TeH XN are as follows:

(1) TeH XN can be identified with the linear space formed by the matrices A ∈ MN (R) P ¯ satisfying k HikHjk(Aik − Ajk) = 0, for any i, j. 0 (2) TH XN consists of those matrices A ∈ MN (R) appearing as Aij = gij(0), where P ¯ i(gik(t)−gjk(t)) g : MN (R)0 → MN (R)0 satisfies k HikHjke = 0 for any i, j. ◦ P ¯ Aik−Ajk (3) TH XN is formed by the matrices A ∈ MN (R) satisfying k HikHjkq = 0, for any i 6= j and any q ∈ T. × (4) TH XN is formed by the matrices A ∈ MN (R) which are of the form Aij = ai + bj, for certain vectors a, b ∈ RN . Proof. All these assertions can be deduced by using basic differential geometry: (1) This result is well-known, the idea being as follows. First, MN (T) is defined by the 2 algebraic relations |Hij| = 1, and with Hij = Xij + iYij we have: 2 2 2 ˙ ˙ d|Hij| = d(Xij + Yij) = 2(XijXij + YijYij) P ˙ ˙ Now since an arbitrary vector ξ ∈ TH MN (C), written as ξ = ij αijXij +βijYij, belongs 2 to TH MN (T) if and only if < ξ, d|Hij| >= 0 for any i, j, we obtain: ( )

X ˙ ˙ TH MN (T) = Aij(YijXij − XijYij) Aij ∈ R ij √ We also know that NUN is deﬁned by the algebraic relations < Hi,Hj >= Nδij, where H1,...,HN are the rows of H. The relations < Hi,Hi >= N being automatic for the matrices H ∈ MN (T), if for i 6= j we let Lij =< Hi,Hj >, then we have: n ˙ o TeH CN = ξ ∈ TH MN (T)| < ξ, Lij >= 0, ∀i 6= j

On the other hand, diﬀerentiating the formula of Lij gives: ˙ X ˙ ˙ ˙ ˙ Lij = (Xik + iYik)(Xjk − iYjk) + (Xjk − iYjk)(Xik + iYik) k

Now if we pick ξ ∈ TH MN (T), written as above in terms of A ∈ MN (R), we obtain: ˙ X ¯ < ξ, Lij >= i HikHjk(Aik − Ajk) k

Thus we have reached to the description of TeH XN in the statement. it igij (t) (2) Pick an arbitrary deformation, and write it as fij(e ) = Hije . Observe ﬁrst that the Hadamard condition corresponds to the equations in the statement, namely:

X ¯ i(gik(t)−gjk(t)) HikHjke = 0 k 56 TEO BANICA

Observe also that by differentiating this formula at t = 0, we obtain: X ¯ 0 0 HikHjk(gik(0) − gjk(0)) = 0 k 0 Thus the matrix Aij = gij(0) belongs indeed to TeH XN , so we obtain in this way a certain map TH XN → TeH XN . In order to check that this map is indeed the correct one, we have to verify that, for any i, j, the tangent vector to our deformation is given by: 0 ˙ ˙ ξij = gij(0)(YijXij − XijYij) But this latter verification is just a one-variable problem. So, by dropping all i, j indices, which is the same as assuming N = 1, we have to check that for any point H ∈ T, written H = X + iY , the tangent vector to the deformation f(eit) = Heig(t) is: ξ = g0(0)(Y X˙ − XY˙ ) But this is clear, because the unit tangent vector at H ∈ T is η = −i(Y X˙ − XY˙ ), and ig(t) 0 0 its coefficient coming from the deformation is (e )|t=0 = −ig (0). (3) Observe first that by taking the derivative at q = 1 of the condition (2) in Propo- sition 4.1, of just by using the condition (3) there with the function ϕ(r) = r, we get: X ¯ HikHjkϕ(Aik − Ajk) = 0 k ◦ Thus we have a map TH XN → TeH XN , and the fact that is map is indeed the correct one comes for instance from the computation in (2), with gij(t) = Aijt. (4) Observe first that the Hadamard matrix condition is satisfied:

X ¯ Aik−Ajk ai−aj X ¯ HikHjkq = q HikHjk = δij k k × As for the fact that TH XN is indeed the space in the statement, this is clear.

Let ZN ⊂ XN be the real algebraic manifold formed by all the dephased N × N complex Hadamard matrices. Observe that we have a quotient map XN → ZN , obtained by dephasing. With this notation, we have the following refinement of (4) above: Proposition 4.5. We have a direct sum decomposition of cones ◦ × ◦ TH XN = TH XN ⊕ TH ZN where at right we have the affine tangent cone to the dephased manifold XN → ZN . ◦ Proof. If we denote by MN (R) the set of matrices having 0 outside the first row and column, we have a direct sum decomposition, as follows: ◦ ◦ ◦ TeH XN = MN (R) ⊕ TeH ZN Now by looking at the affine cones, and using Theorem 4.4, this gives the result. HADAMARD MATRICES 57

Summarizing, we have so far a number of theoretical results about the tangent cones TH XN that we are interested in, and their versions coming from√ the trivial and affine deformations, and from the intersection formula XN = MN (T) ∩ NUN as well. In practice now, passed a few special cases where all these cones collapse to the trivial × cone TN XN , which by Proposition 4.5 means that the image of H ∈ XN must be isolated in the dephased manifold XN → ZN , things are quite difficult to compute. However, as a concrete numerical invariant arising from all this, which can be effectively computed in many cases of interest, we have, following [88]: Definition 4.6. The real dimension d(H) of the enveloping tangent space √ TeH XN = TH MN (T) ∩ TH NUN is called undephased defect of a complex Hadamard matrix H ∈ XN . In view of Proposition 4.5, it is sometimes convenient to replace d(H) by the related quantity d0(H) = d(H) − 2N + 1, called dephased defect of H. See [88]. In what follows we will rather use d(H) as defined above, and simply call it “defect” of H. We already know, from Theorem 4.4, what is the precise geometric meaning of the defect, and how to compute it. Let us record again these results, that we will use many times in what follows, in a slightly different form, closer to the spirit of [88]: Theorem 4.7. The defect d(H) is the real dimension of the linear space ( )

X ¯ TeH XN = A ∈ MN (R) HikHjk(Aik − Ajk) = 0, ∀i, j k

q Aij and the elements of this space are those making Hij = Hijq Hadamard at order 1. Proof. Here the first assertion is something that we already know, from Theorem 4.4 (1), and the second assertion follows either from Theorem 4.4 and its proof, or directly from the definition of the enveloping tangent space TeH XN , as used in Definition 4.6. Here are a few basic properties of the defect:

Proposition 4.8. Let H ∈ XN be a complex Hadamard matrix. (1) If H ' He then d(H) = d(He). (2) We have 2N − 1 ≤ d(H) ≤ N 2. (3) If d(H) = 2N − 1, the image of H in the dephased manifold XN → ZN is isolated. Proof. All these results are elementary, the proof being as follows: (1) If we let Kij = aibjHij with |ai| = |bj| = 1 be a trivial deformation of our matrix H, the equations for the enveloping tangent space for K are: X ¯ ¯ aibkHika¯jbkHjk(Aik − Ajk) = 0 k 58 TEO BANICA

By simplifying we obtain the equations for H, so d(H) is invariant under trivial deformations. Since d(H) is invariant as well by permuting rows or columns, we are done. × × (2) Consider the inclusions TH XN ⊂ TH XN ⊂ TeH XN . Since dim(TH XN ) = 2N − 1, the inequality at left holds indeed. As for the inequality at right, this is clear. × (3) If d(H) = 2N − 1 then TH XN = TH XN , so any deformation of H is trivial. Thus the image of H in the quotient manifold XN → ZN is indeed isolated, as stated. Let us discuss now the computation of the defect for the most basic examples of complex Hadamard matrices that we know, namely the real ones, and the Fourier ones. In order to deal with the real case, it is convenient to modify the general formula from Theorem 4.7 above, via a change of variables, as follows: Proposition 4.9. We have a linear space isomorphism as follows, n o ∗ ¯ TeH XN ' E ∈ MN (C) E = E , (EH)ijHij ∈ R, ∀i, j the correspondences A → E and E → A being given by the formulae X ¯ ¯ Eij = HikHjkAik ,Aij = (EH)ijHij k with A ∈ TeH XN being the usual components, from Theorem 4.7 above. ∗ Proof. Given a matrix A ∈ MN (C), if we set Rij = AijHij and E = RH , the correspondence A → R → E is then bijective onto MN (C), and we have: X ¯ Eij = HikHjkAik k In terms of these new variables, the equations in Theorem 4.7 become: ¯ Eij = Eji Thus, when taking into account these conditions, we are simply left with the conditions ¯ Aij ∈ R. But these correspond to the conditions (EH)ijHij ∈ R, as claimed. With the above result in hand, we can now compute the defect of the real Hadamard matrices. The result here, from [86], is as follows:

Theorem 4.10. For any real Hadamard matrix H ∈ MN (±1) we have

symm TeH XN ' MN (R) and so the corresponding defect is d(H) = N(N + 1)/2. ¯ Proof. We use Proposition 4.9. Since H is now real the condition (EH)ijHij ∈ R there simply tells us that E must be real, and this gives the result. HADAMARD MATRICES 59

We should mention that the above result, as well as the whole basic theory of the tangent cones and defect, can be extended in a quite straightforward way to the case of the partial Hadamard matrices [22]. We will be back to this, later on.

Let us discuss now the computation of the defect of the Fourier matrix FG. The main idea here goes back to [62], with some supplementary contributions from [72], the main formula, in the cyclic group case, was obtained in [88], the extension to the general case was done in [5], and the corresponding deformations were studied in [73]. As a ﬁrst result on this subject, we have, following [88]:

Theorem 4.11. For a Fourier matrix F = FG, the matrices A ∈ TeF XN , with N = |G|, ∗ are those of the form A = PF , with P ∈ MN (C) satisfying ¯ Pij = Pi+j,j = Pi,−j where the indices i, j are by deﬁnition taken in the group G. Proof. We use the system of equations in Theorem 4.7, namely: X ¯ FikFjk(Aik − Ajk) = 0 k

By decomposing our ﬁnite abelian group as G = ZN1 × ... × ZNr we can assume 2πi/k F = FN1 ⊗ ... ⊗ FNr , so that with wk = e we have:

i1j1 irjr Fi1...ir,j1...jr = (wN1 ) ... (wNr ) 2πi/N With N = N1 ...Nr and w = e , we obtain: i1j1 +...+ irjr N N1 Nr Fi1...ir,j1...jr = w Thus the matrix of our system is given by: (i1−j1)k1 +...+ (ir−jr)kr N ¯ N1 Nr Fi1...ir,k1...kr Fj1...jr,k1...kr = w Now by plugging in a multi-indexed matrix A, our system becomes: X (i1−j1)k1 +...+ (ir−jr)kr N N1 Nr w (Ai1...ir,k1...kr − Aj1...jr,k1...kr ) = 0

k1...kr Now observe that in the above formula we have in fact two matrix multiplications, so our system can be simply written as:

(AF )i1...ir,i1−j1...ir−jr − (AF )j1...jr,i1−j1...ir−jr = 0 Now recall that our indices have a “cyclic” meaning, so they belong in fact to the group G. So, with P = AF , and by using multi-indices, our system is simply:

Pi,i−j = Pj,i−j

With i = I + J, j = I we obtain the condition PI+J,J = PIJ in the statement. 60 TEO BANICA

∗ ˜ ¯ In addition, A = PF must be a real matrix. But, if we set Pij = Pi,−j, we have: X (PF ∗) = P¯ F i1...ir,j1...jr i1...ir,k1...kr j1...jr,k1...kr k1...kr X ˜ ∗ = Pi1...ir,−k1...−kr (F )−k1...−kr,j1...jr

k1...kr ˜ ∗ = (PF )i1...ir,j1...jr Thus we have PF ∗ = PF˜ ∗, so the fact that the matrix PF ∗ is real, which means by deﬁnition that we have PF ∗ = PF ∗, can be reformulated as PF˜ ∗ = PF ∗, and hence as ˜ ¯ P = P . So, we obtain the conditions Pij = Pi,−j in the statement. We can now compute the defect, and we are led to the following formula:

Theorem 4.12. The defect of a Fourier matrix FG is given by X |G| d(F ) = G ord(g) g∈G and equals as well the number of 1 entries of the matrix FG. ∗ Proof. According to the formula A = PF from Theorem 4.11, the defect d(FG) is the dimension of the real vector space formed by the matrices P ∈ MN (C) satisfying: ¯ Pij = Pi+j,j = Pi,−j Here, and in what follows, the various indices i, j, . . . will be taken in G. Now the point is that, in terms of the columns of our matrix P , the above conditions are: (1) The entries of the j-th column of P , say C, must satisfy Ci = Ci+j. (2) The (−j)-th column of P must be conjugate to the j-th column of P . Thus, in order to count the above matrices P , we can basically ﬁll the columns one by one, by taking into account the above conditions. In order to do so, consider the subgroup G2 = {j ∈ G|2j = 0}, and then write G as a disjoint union, as follows:

G = G2 t X t (−X)

With this notation, the algorithm is as follows. First, for any j ∈ G2 we must fill the j-th column of P with real numbers, according to the periodicity rule Ci = Ci+j. Then, for any j ∈ X we must fill the j-th column of P with complex numbers, according to the same periodicity rule Ci = Ci+j. And finally, once this is done, for any j ∈ X we just have to set the (−j)-th column of P to be the conjugate of the j-th column. So, let us compute the number of choices for filling these columns. Our claim is that, when uniformly distributing the choices for the j-th and (−j)-th columns, for j∈ / G2, there are exactly [G :< j >] choices for the j-th column, for any j. Indeed: (1) For the j-th column with j ∈ G2 we must simply pick N real numbers subject to the condition Ci = Ci+j for any i, so we have indeed [G :< j >] such choices. HADAMARD MATRICES 61

(2) For ﬁlling the j-th and (−j)-th column, with j∈ / G2, we must pick N complex numbers subject to the condition Ci = Ci+j for any i. Now since there are [G :< j >] choices for these numbers, so a total of 2[G :< j >] choices for their real and imaginary parts, on average over j, −j we have [G :< j >] choices, and we are done again. Summarizing, the dimension of the vector space formed by the matrices P , which is equal to the number of choices for the real and imaginary parts of the entries of P , is: X d(FG) = [G :< j >] j∈G But this is exactly the number in the statement. Regarding now the second assertion, according to the abstract deﬁnition of the Fourier matrix FG, from Theorem 2.8 above, the number of 1 entries of FG is given by: n o

#(1 ∈ FG) = # (g, χ) ∈ G × Gb χ(g) = 1 n o X = # χ ∈ Gb χ(g) = 1 g∈G X |G| = ord(g) g∈G

Thus, the second assertion follows from the ﬁrst one.

Let us ﬁnish now the work, and explicitely compute the defect of FG. It is convenient to consider the following quantity, which behaves better: X 1 δ(G) = ord(g) g∈G

a As a ﬁrst example, consider a cyclic group G = ZN , with N = p power of a prime. The count here is very simple, over sets of elements having a given order:

−1 2 −2 a a−1 −1 δ(Zpa ) = 1 + (p − 1)p + (p − p)p + ... + (p − p )p a = 1 + a − p In order to extend this kind of count to the general abelian case, we use two ingredients. First is the following result, which splits the computation over isotypic components: Proposition 4.13. For any ﬁnite groups G, H we have: δ(G × H) ≥ δ(G)δ(H) In addition, if (|G|, |H|) = 1, we have equality. 62 TEO BANICA

Proof. Indeed, we have the following estimate: X 1 δ(G × H) = ord(g, h) gh X 1 = [ord(g), ord(h)] gh X 1 ≥ ord(g) · ord(h) gh = δ(G)δ(H) Now in the case (|G|, |H|) = 1, the least common multiple appearing on the right becomes a product, [ord(g), ord(h)] = ord(g) · ord(h), so we have equality. We deduce from this that we have the following result: Proposition 4.14. For a ﬁnite abelian group G we have Y δ(G) = δ(Gp) p where Gp with G = ×pGp are the isotypic components of G.

Proof. This is clear from Proposition 4.13, because the order of Gp is a power of p. As an illustration, we can recover in this way the defect computation in [89]:

Theorem 4.15. The defect of a usual Fourier matrix FN is given by s Y ai d(F ) = N 1 + a − N i p i=1 i

a1 as where N = p1 . . . ps is the decomposition of N into prime factors.

Proof. The underlying group here is the cyclic group G = ZN , whose isotypic components are the cyclic groups Gp = ai . By applying now Proposition 4.14, and by using the i Zpi computation for cyclic p-groups performed before Proposition 4.13, we obtain: s Y −1 d(FN ) = N 1 + pi (pi − 1)ai i=1 But this is exactly the formula in the statement.

Now back to the general case, where we have an arbitrary Fourier matrix FG, we will need, as a second ingredient for our computation, the following result: HADAMARD MATRICES 63

Proposition 4.16. For the p-groups, the quantities n o k ck = # g ∈ G ord(g) ≤ p are multiplicative, in the sense that ck(G × H) = ck(G)ck(H). Proof. Indeed, for a product of p-groups we have: n o k ck(G × H) = # (g, h) ord(g, h) ≤ p n o k k = # (g, h) ord(g) ≤ p , ord(h) ≤ p n o n o k k = # g ord(g) ≤ p # h ord(h) ≤ p

We recognize at right ck(G)ck(H), and we are done. Let us compute now δ in the general isotypic case:

Proposition 4.17. For G = Zpa1 × ... × Zpar with a1 ≤ a2 ≤ ... ≤ ar we have r X (r−k)ak−1+(a1+...+ak−1)−1 r−k+1 δ(G) = 1 + p (p − 1)[ak − ak−1]pr−k k=1 2 a−1 with the convention a0 = 0, and by using the notation [a]q = 1 + q + q + ... + q .

Proof. First, in terms of the numbers ck, we have: X ck − ck−1 δ(G) = 1 + pk k≥1 min(k,a) In the case of a cyclic group G = Zpa we have ck = p . Thus, in the general isotypic case G = Zpa1 × ... × Zpar we have:

min(k,a1) min(k,ar) min(k,a1)+...+min(k,ar) ck = p . . . p = p Now observe that the exponent on the right is a piecewise linear function of k. More precisely, by assuming a1 ≤ a2 ≤ ... ≤ ar as in the statement, the exponent is linear on each of the intervals [0, a1], [a1, a2],..., [ar−1, ar]. So, the quantity δ(G) to be computed will be 1 plus the sum of 2r geometric progressions, 2 for each interval. In practice now, the numbers ck are as follows:

r 2r ra1 c0 = 1, c1 = p , c2 = p , . . . , ca1 = p ,

a1+(r−1)(a1+1) a1+(r−1)(a1+2) a1+(r−1)a2 ca1+1 = p , ca1+2 = p , . . . , ca2 = p ,

a1+a2+(r−2)(a2+1) a1+a2+(r−2)(a2+2) a1+a2+(r−2)a3 ca2+1 = p , ca2+2 = p , . . . , ca3 = p , ......

a1+...+ar−1+(ar−1+1) a1+...+ar−1+(ar−1+2) a1+...+ar car−1+1 = p , car−1+2 = p , . . . , car = p 64 TEO BANICA

Now by separating the positive and negative terms in the above formula of δ(G), we have indeed 2r geometric progressions to be summed, as follows:

δ(G) = 1 + (pr−1 + p2r−2 + p3r−3 + ... + pa1r−a1 ) −(p−1 + pr−2 + p2r−3 + ... + p(a1−1)r−a1 ) +(p(r−1)(a1+1)−1 + p(r−1)(a1+2)−2 + ... + pa1+(r−2)a2 ) −(pa1r−a1−1 + p(r−1)(a1+1)−2 + ... + pa1+(r−1)(a2−1)−a2 ) + ... +(pa1+...+ar−1 + pa1+...+ar−1 + ... + pa1+...+ar−1 ) −(pa1+...+ar−1−1 + pa1+...+ar−1−1 + ... + pa1+...+ar−1−1)

Now by performing all the sums, we obtain:

p(r−1)a1 − 1 δ(G) = 1 + p−1(pr − 1) pr−1 − 1 p(r−2)(a2−a1) − 1 +p(r−2)a1+(a1−1)(pr−1 − 1) pr−2 − 1 p(r−3)(a3−a2) − 1 +p(r−3)a2+(a1+a2−1)(pr−2 − 1) pr−3 − 1 + ...

a1+...+ar−1−1 +p (p − 1)(ar − ar−1)

By looking now at the general term, we get the formula in the statement.

Let us go back now to the general defect formula in Theorem 4.12. By putting it together with the various results above, we obtain:

Theorem 4.18. For a ﬁnite abelian group G, decomposed as G = ×pGp, we have

r ! Y X (r−k)ak−1+(a1+...+ak−1)−1 r−k+1 d(FG) = |G| 1 + p (p − 1)[ak − ak−1]pr−k p k=1 where a0 = 0 and a1 ≤ a2 ≤ ... ≤ ar are such that Gp = Zpa1 × ... × Zpar .

Proof. Indeed, we know from Theorem 4.12 that we have d(FG) = |G|δ(G), and the result follows from Proposition 4.14 and Proposition 4.17. HADAMARD MATRICES 65

As a ﬁrst illustration, we can recover in this way the formula in Theorem 4.15. Assuming a1 as that N = p1 . . . ps is the decomposition of N into prime factors, we have: s Y −1 d(FN ) = N 1 + pi (pi − 1)ai i=1 s Y ai = N 1 + a − i p i=1 i

As a second illustration, for the group G = Zpa1 × Zpa2 with a1 ≤ a2 we obtain:

a1+a2 −1 2 a1−1 d(FG) = p (1 + p (p − 1)[a1]p + p (p − 1)(a2 − a1)) pa1 − 1 = pa1+a2−1(p + (p2 − 1) + pa1 (p − 1)(a − a )) p − 1 2 1 a1+a2−1 a1 a1 = p (p + (p + 1)(p − 1) + p (p − 1)(a2 − a1)) Finally, let us mention that for general non-abelian groups, there does not seem to be any reasonable algebraic formula for the quantity δ(G). As an example, consider the dihedral group DN , consisting of N symmetries and N rotations. We have: N δ(D ) = + δ( ) N 2 ZN Q −1 Now by remembering the formula for ZN , namely δ(ZN ) = (1 + pi (pi − 1)ai), it is quite clear that the N/2 factor can not be incorporated in any nice way. See [5].

Let us prove now, following the paper of Nicoara and White [73], that for the Fourier matrices the defect is “attained”, in the sense that the deformations at order 0 are true deformations, at order ∞. This is something quite surprising, and non-trivial. Let us begin with some generalities. We ﬁrst recall that we have:

Proposition 4.19. The unitary matrices U ∈ UN around 1 are of the form U = eA with A being an antihermitian matrix, A = −A∗, around 0. Proof. This is something well-known. Indeed, assuming that a matrix A is antihermitian, A = −A∗, the matrix U = eA follows to be unitary: UU ∗ = eA(eA)∗ = eAeA∗ = eAe−A = 1 As for the converse, this follows either by using a dimension argument, which shows that the space of antihermitian matrices is the correct one, or by diagonalizing U. Now back to the Hadamard matrices, we will need to rewrite a part of the basic theory of the defect, using deformations of type t → UtH. First, we have: 66 TEO BANICA

Theorem 4.20. Assume that H ∈ MN (C) is Hadamard, let A ∈ MN (C) be antihermitian, and consider the matrix UH, where U = etA, with t ∈ R. P ¯ tA −tA (1) UH is Hadamard when | rs HrqHsq(e )pr(e )sp| = 1, for any p, q. (2) UH is Hadamard at order 0 when |(AH)pq| = 1, for any p, q. Proof. We already know that UH is unitary, so we must ﬁnd the conditions which guar- antee that we have UH ∈ MN (T), in general, and then at order 0. (1) We have the following computation, valid for any unitary U: 2 |(UH)pq| = (UH)pq(UH)pq ∗ ∗ = (UH)pq(H U )qp X ∗ ∗ = UprHrq(H )qs(U )sp rs X ¯ ¯ = HrqHsqUprUps rs Now with U = etA as in the statement, we obtain: tA 2 X ¯ tA −tA |(e H)pq| = HrqHsq(e )pr(e )sp rs Thus, we are led to the conclusion in the statement. (2) The derivative of the function computed above, taken at 0, is as follows: tA 2 ∂|(e H)pq| X ¯ tA tA = HrqHsq(e A)pr(−e A)sp |t=0 ∂t |t=0 rs X ¯ = HrqHsqApr(−A)sp rs X X ∗ ∗ = AprHrq (H )qs(A )sp r s ∗ ∗ = (AH)pq(H A )qp 2 = |(AH)pq| Thus, we obtain the conclusion in the statement. In the Fourier matrix case we can go beyond this, and we have:

Proposition 4.21. Given a Fourier matrix FG ∈ MG(C), and an antihermitian matrix tA A ∈ MG(C), the matrix UFG, where U = e with t ∈ R, is Hadamard when

X X tm X m X Ak (−A)l = δ m! l p,s+n sp n0 s m k+l=m s for any p, with the indices being k, l, m ∈ N, and n, p, s ∈ G. HADAMARD MATRICES 67

Proof. According to the formula in the proof of Theorem 4.20 (1), we have:

2 X tA −tA |(UFG)pq| = (FG)rq(FG)sq(e )pr(e )sp rs X tA −tA = < r, q >< −s, q > (e )pr(e )sp rs X tA −tA = < r − s, q > (e )pr(e )sp rs By setting n = r − s, can write this formula in the following way:

2 X tA −tA |(UFG)pq| = < n, q > (e )p,s+n(e )sp ns X X tA −tA = < n, q > (e )p,s+n(e )sp n s Since this quantity must be 1 for any q, we must have:

X tA −tA (e )p,s+n(e )sp = δn0 s On the other hand, we have the following computation:

X tA −tA (e )p,s+n(e )sp s k l X X (tA)p,s+n (−tA)sp = · k! l! s kl X X 1 X = (tA)k (−tA)l k!l! p,s+n sp s kl s X X tk+l X = Ak (−A)l k!l! p,s+n sp s kl s X X X 1 X = tm Ak (−A)l k!l! p,s+n sp s m k+l=m s X X tm X m X = Ak (−A)l m! l p,s+n sp s m k+l=m s

Thus, we obtain the conclusion in the statement.

Following [73], let us construct now the deformations. The result is as follows: 68 TEO BANICA

Theorem 4.22. Let G be a ﬁnite abelian group, and for any g, h ∈ G, let us set: ( 1 if ∃k ∈ , p = hkg, q = hk+1g B = N pq 0 otherwise When (g, h) ∈ G2 range in suitable cosets, the unitary matrices it(B+Bt) t(B−Bt) e FG , e FG are both Hadamard, and make the defect of FG to be attained. Proof. The proof of this result, from [73], is quite long and technical, based on the Fourier computation from Proposition 4.21 above, the idea being as follows: (1) First of all, an elementary algebraic study shows that when (g, h) ∈ G2 range in some suitable cosets, coming from the proof of Theorem 4.12, the various matrices B = Bgh constructed above are distinct, the matrices A = i(B + Bt) and A0 = B − Bt are linearly independent, and the number of such matrices equals the defect of FG. (2) It is also standard to check that each B = (Bpq) is a partial isometry, and that Bk,B∗k are given by simple formulae. With this ingredients in hand, the Hadamard property follows from the Fourier computation from the proof of Proposition 4.21. Indeed, we can compute the exponentials there, and eventually use the binomial formula. (3) Finally, the matrices in the statement can be shown to be non-equivalent, and this is something more technical, for which we refer to [73]. With this last ingredient in hand, a comparison with Theorem 4.12 shows that the defect of FG is indeed attained, in the sense that all order 0 deformations are actually true deformations. See [73]. The above result is something quite surprising, which came a long time after the original defect paper [88], and even more time after the early computations in [62]. Let us also mention that [73] was written in terms of subfactor-theoretic commuting squares, with a larger class of squares actually under investigation. We will discuss the relation between Hadamard matrices and commuting squares in section 11 below. HADAMARD MATRICES 69

5. Special matrices We have seen in the previous section that the defect theory from [88] can be successfully applied to the real Hadamard matrices, and to the Fourier matrices. We discuss here a number of more specialized aspects, regarding the tensor products, the Dit¸˘adeformations of such tensor products, the Butson and the regular matrices, the master Hadamard matrices, the McNulty-Weigert matrices, and ﬁnally the partial Hadamard matrix case, following [2], [5], [6], [22], [69], [88], [89].

Let us begin with some generalities. The standard defect equations are those in Theo- rem 4.7, naturally coming from the computations in the proof of Theorem 4.4: ( ) X d(H) = dim A ∈ M ( ) H H¯ (A − A ) = 0, ∀i, j R N R ik jk ik jk k However, we have seen that for various concrete questions, some manipulations on these equations are needed. To be more precise, the study in the real case was based on the P ¯ transformation Eij = k HikHjkAik from Proposition 4.9, the study in the Fourier matrix case was based on the transformation P = AFG from Theorem 4.11, and the ﬁne study in the Fourier matrix case was based on the t → etAH method from Proposition 4.21. In short, each type of complex Hadamard matrix seems to require its own general theory, and defect manipulations, and there is no way of escaping from this.

In view of this phenomenon, let us first present some further manipulations on the defect equations, which are all quite natural, and potentially useful. First, we have: Proposition 5.1. The defect d(H) is the corank of the matrix  Re(H H¯ ) if i < j  ib jb ¯ Yij,ab = (δia − δja) Im(HibHjb) if i > j ∗ if i = j where ∗ can be any quantity, its coefficient being 0 anyway. Proof. The matrix of the system defining the enveloping tangent space is: ¯ Xij,ab = (δia − δja)HibHjb

However, since we are only looking for real solutions A ∈ MN (R), we have to take into account the real and imaginary parts. But this is not a problem, because the (ij) equation coincides with the (ji) one, that we can cut. More precisely, if we set Y as above, then we obtain precisely the original system. Thus the defect of H is the corank of Y . 70 TEO BANICA

As an illustration, for the Fourier matrix FN we have the following formula, where e(i, j) ∈ {−1, 0, 1} is negative if i < j, null for i = j, and positive for i > j: 1 Y = (δ − δ )(w(i−j)b + e(i, j)w(j−i)b) ij,ab 2 ia ja

Observe in particular that for the Fourier matrix F2 we have:  0 0 0 0   1 −1 −1 1  Y =   −1 1 1 −1 0 0 0 0

Here the corank is 3, but, unfortunately, this cannot be seen on the characteristic polynomial, which is P (λ) = λ4. The problem is that our matrix, and more precisely its middle 2 × 2 block, is not diagonalizable. This phenomenon seems to hold in general.

A second possible manipulation, which is of interest in connection with quantum groups and subfactor theory, concerns the reformulation of the defect in terms of the proﬁle matrix of H. Indeed, it is known that both the quantum group and subfactor associated to H depend only on this proﬁle matrix, and this will be explained in sections 10-11 below. We do not have an answer here, but our conjecture is as follows:

Conjecture 5.2. The proﬁle matrix of H, namely

jb X ¯ ¯ Mia = HikHjkHakHbk k determines the enveloping tangent space TeH XN , or at least the defect d(H). All this is of course related to the general question on whether the associated quantum group or subfactor can “see” the defect, via various representation theory invariants. For a number of further speculations on all this, and on some related glow questions as well, in relation with the general theory in [45], [46], we refer to [5], [6], [8].

Let us get back now to our original goal here, namely computing the defect for various classes of special matrices. For the tensor products, we have the following result:

Proposition 5.3. For a tensor product L = H ⊗ K we have

d(L) ≥ d(H)d(K) coming from an inclusion of linear spaces TeH XM ⊗ TeK XN ⊂ TeLXMN . HADAMARD MATRICES 71

Proof. For a matrix A = B ⊗ C, we have the following formulae: X X ¯ X ¯ (H ⊗ K)ia,kc(H ⊗ K)jb,kcAia,kc = HikHjkBik KacKbcCac kc k c X X ¯ X ¯ (H ⊗ K)ia,kc(H ⊗ K)jb,kcAjb,kc = HikHjkBjk KacKbcCbc kc k c

Now by assuming B ∈ TeH XM and C ∈ TeK XN , the two quantities on the right are equal. Thus we have indeed A ∈ TeLXMN , and we are done. Observe that we do not have equality in the tensor product estimate, even in very simple cases. For instance if we consider two Fourier matrices F2, we obtain: 2 d(F2 ⊗ F2) = 10 > 9 = d(F2) In fact, besides the isotypic decomposition results from section 4 above, valid for the Fourier matrices, there does not seem to be anything conceptual on this subject. We will be back to this, however, in Theorem 5.6 below, with a slight advance on all this.

Let us discuss now the Dit¸˘adeformations. Here the study is even more diﬃcult, and we basically have just one result, when the deformation matrix is as follows:

Deﬁnition 5.4. A rectangular matrix Q ∈ MM×N (T) is called “dephased and elsewhere generic” if the entries on its ﬁrst row and column are all equal to 1, and the remaining (M − 1)(N − 1) entries are algebrically independent over Q. Here the last condition takes of course into account the fact that the entries of Q themselves have modulus 1, the independence assumption being modulo this fact. With this convention made, we have the following result:

Theorem 5.5. If H ∈ XM ,K ∈ XN are dephased, of Butson type, and Q ∈ MM×N (T) is dephased and elsewhere generic, then A = (Aia,kc) belongs to TeH⊗QK XMN iﬀ

ij ij ij ji ii Aac = Abc ,Aac = Aac , (Axy)xy ∈ TeK XN ij P ¯ hold for any a, b, c and i 6= j, where Aac = k HikHjkAia,kc. Proof. Consider the system for the enveloping tangent space, namely: X (H ⊗Q K)ia,kc(H ⊗Q K)jb,kc(Aia,kc − Ajb,kc) = 0 kc

We have (H ⊗Q K)ia,jb = qibHijKab, and so our system is: X ¯ X ¯ qicq¯jcKacKbc HikHjk(Aia,kc − Ajb,kc) = 0 c k 72 TEO BANICA

ij P ¯ Consider now the variables Aac = k HikHjkAia,kc in the statement. We have: ij X ¯ X ¯ Aac = HikHjkAia,kc = HjkHikAia,kc k k Thus, in terms of these variables, our system becomes simply: X ¯ ij ji qicq¯jcKacKbc(Aac − Abc) = 0 c More precisely, the above equations must hold for any i, j, a, b. By distinguishing now two cases, depending on whether i, j are equal or not, the situation is as follows: (1) Case i 6= j. In this case, let us look at the row vector of parameters, namely:

(qicq¯jc)c = (1, qi1q¯j1, . . . , qiM q¯jM ) Now since Q was assumed to be dephased and elsewhere generic, and because of our assumption i 6= j, the entries of the above vector are linearly independent over Q¯ . But, since by linear algebra we can restrict attention to the computation of the solutions over ¯ ij ji Q, the i 6= j part of our system simply becomes Aac = Abc, for any a, b, c and any i 6= j. Now by making now a, b, c vary, we are led to the following equations:

ij ij ij ji Aac = Abc,Aac = Aac, ∀a, b, c, i 6= j (2) Case i = j. In this case the parameters cancel, and our equations become:

X ¯ ii ii KacKbc(Aac − Abc) = 0, ∀a, b, c, i c ii P On the other hand, we have Aac = k Aia,kc, and so our equations become: X ¯ ii ii KacKbc(Aac − Abc) = 0, ∀a, b, c, i c

But these are precisely the equations for the space TeK XN , and we are done. Let us go back now to the usual tensor product situation, and look at the aﬃne cones. ◦ The problem here is that of ﬁnding the biggest subcone of TH⊗K XMN , obtained by gluing ◦ ◦ TH XM ,TK XN . Our answer here, which takes into account the two “semi-trivial” cones coming from the left and right Dit¸˘adeformations, is as follows:

◦ ◦ Theorem 5.6. The cones TH XM = {B} and TK XN = {C} glue via the formulae

Aia,jb = λBij + ψjCab + Xia + Yjb + Faj

Aia,jb = φbBij + µCab + Xia + Yjb + Eib ◦ producing in this way two subcones of the aﬃne cone TH⊗K XMN = {A}. HADAMARD MATRICES 73

Proof. Indeed, the idea is that Xia,Yjb are the trivial parameters, and that Eib,Faj are the Dit¸˘aparameters. In order to prove the result, we use the criterion in Theorem 4.4 (3) above. So, given a matrix A = (Aia,jb), consider the following quantity:

X ¯ ¯ Aia,kc−Ajb,kc P = HikHjkKacKbcq kc Let us prove now the ﬁrst statement, namely that for any choice of matrices B ∈ ◦ ◦ TH XM ,C ∈ TH XN and of parameters λ, ψj,Xia,Yjb,Faj, the ﬁrst matrix A = (Aia,jb) ◦ constructed in the statement belongs indeed to TH⊗K XMN . We have:

Aia,kc = λBik + ψkCac + Xia + Ykc + Fak

Ajb,kc = λBjk + ψkCbc + Xjb + Ykc + Fbk Now by substracting, we obtain:

Aia,kc − Ajb,kc = λ(Bik − Bjk) + ψk(Cac − Cbc) + (Xia − Xjb) + (Fak − Fbk) It follows that the above quantity P is given by:

X ¯ ¯ λ(Bik−Bjk)+ψk(Cac−Cbc)+(Xia−Xjb)+(Fak−Fbk) P = HikHjkKacKbcq kc

Xia−Xjb X ¯ Fak−Fbk λ(Bik−Bjk) X ¯ ψk Cac−Cbc = q HikHjkq q KacKbc(q ) k c

Xia−Xja X ¯ λ Bik−Bjk = δabq HikHjk(q ) k = δabδij ◦ Thus Theorem 4.4 (3) applies and tells us that we have A ∈ TH⊗K XMN , as claimed. In the second case now, the proof is similar. First, we have:

Aia,kc = φcBik + µCac + Xia + Ykc + Eic

Ajb,kc = φcBjk + µCbc + Xjb + Ykc + Ejc Thus by substracting, we obtain:

Aia,kc − Ajb,kc = φc(Bik − Bjk) + µ(Cac − Cbc) + (Xia − Xjb) + (Eic − Ejc) It follows that the above quantity P is given by:

X ¯ ¯ φc(Bik−Bjk)+µ(Cac−Cbc)+(Xia−Xjb)+(Eic−Ejc) P = HikHjkKacKbcq kc

Xia−Xjb X ¯ Eic−Ejc µ(Cac−Cbc) X ¯ φc Bik−Bjk = q KacKbcq q HikHjk(q ) c k

Xia−Xib X ¯ µ Cac−Cbc = δijq KacKbc(q ) = δijδab c Thus Theorem 4.4 (3) applies again, and gives the result. 74 TEO BANICA

We believe Theorem 5.6 above to be “optimal”, in the sense that nothing more can be ◦ said about the aﬃne tangent space TH⊗K XMN , in the general case. See [6]. Let us discuss now some rationality questions, in relation with:

Deﬁnition 5.7. The rational enveloping tangent space at H ∈ XN is

[TeH XN ]Q = TeH XN ∩ MN (Q) and the dimension dQ(H) of this space is called rational defect of H. Observe that the first notion can be extended to all the tangent cones at H, and by using an arbitrary field K ⊂ C instead of Q. Indeed, we can set: ∗ ∗ [TH XN ]K = TH XN ∩ MN (K) However, in what follows we will be interested only in the objects constructed in Defi- nition 5.7. It follows from definitions that dQ(H) ≤ d(H), and we have:

Conjecture 5.8. For a regular Hadamard matrix H ∈ MN (C) we have

TeH CN = C · [TeH CN ]Q and so the defect equals the rational defect, dQ(H) = d(H).

For the usual Fourier matrices FN , this deﬁnitely holds at N = p prime, because the minimal polynomial of w over Q is simply P = 1 + w + ... + wp−1. The case N = p2 has also a simple solution, coming from the fact that all the p × p blocks of our matrix A can be shown to coincide. In general, this method should probably work for N = pk, or even for any N ∈ N. So, our conjecture would be that this holds indeed for the Fourier matrices, usual or general, and more generally for the Butson matrices, and even more generally, modulo Conjecture 3.22 above, for the regular matrices. See [6].

Let us discuss now defect computations for a very interesting class of Hadamard matrices, namely the “master” ones, introduced in [2]: Deﬁnition 5.9. A master Hadamard matrix is an Hadamard matrix of the form

nj Hij = λi

P nj with λi ∈ T, nj ∈ R. The associated “master function” is f(z) = j z .

imi iminj Observe that with λi = e we have Hij = e . The basic example of such a matrix is the Fourier matrix FN , having master function as follows: zN − 1 f(z) = z − 1 Observe that, in terms of f, the Hadamard condition on H is simply: λi f = Nδij λj HADAMARD MATRICES 75

These matrices were introduced in [2], the motivating remark there being the fact that the following operator deﬁnes a representation of the Temperley-Lieb algebra [92]:

X ni−nj R = eij ⊗ Λ ij At the level of examples, the ﬁrst observation, from [2], is that the standard 4 × 4 complex Hadamard matrices are, with 2 exceptions, master Hadamard matrices: Proposition 5.10. The following complex Hadamard matrix, with |q| = 1, 1 1 1 1  q 1 −1 1 −1 F =   2,2 1 q −1 −q 1 −q −1 q is a master Hadamard matrix, for any q 6= ±1.

Proof. We use the exponentiation convention (eit)r = eitr for t ∈ [0, 2π) and r ∈ R. Since q2 6= 1, we can ﬁnd k ∈ R such that q2k = −1, and so our matrix becomes:  10 11 12k 12k+1  0 1 2k 2k+1 q (−1) (−1) (−1) (−1)  F =  0 1 2k 2k+1  2,2  q q q q  (−q)0 (−q)1 (−q)2k (−q)2k+1

x y z t Now if we pick λ 6= 1 and write 1 = λ , −1 = λ , q = λ , −q = λ , we are done. We have the following generalization of Proposition 5.10, once again from [2]:

Theorem 5.11. FM ⊗Q FN is master Hadamard, for any Q ∈ MM×N (T) of the form

i(Npb+b) Qib = q 2πi/MNk where q = e with k ∈ N, and p0, . . . , pN−1 ∈ R. Proof. The main construction in [2] is, in terms of master functions, as follows: Nk f(z) = fM (z )fN (z) Here k ∈ N, and the functions on the right are by deﬁnition as follows:

X Mri+i X Npa+a fM (z) = z , fN (z) = z i a i a 2πi/N Nk We use the eigenvalues λia = q w , where w = e , and where q = ν, where M Nk ν = 1. Observe that, according to f(z) = fM (z )fN (z), the exponents are:

njb = Nk(Mrj + j) + Npb + b 76 TEO BANICA

Thus the associated master Hadamard matrix is given by:

i a Nk(Mrj +j)+Npb+b Hia,jb = (q w ) = νijqi(Npb+b)wa(Npb+b) = νijwabqi(Npb+b)

ij ab i(Npb+b) Now since (FM ⊗ FN )ia,jb = ν w , we get H = FM ⊗Q FN with Qib = q , as claimed. Observe that Q itself is a “master matrix”, because the indices split. In view of the above examples, and of the lack of other known examples of master Hadamard matrices, he following conjecture was made in [2]: Conjecture 5.12 (Master Hadamard Conjecture). The master Hadamard matrices appear as Dit¸˘adeformations of FN . There is a relation here with the notions of defect and isolation, that we would like to discuss now. First, we have the following defect computation: Theorem 5.13. The defect of a master Hadamard matrix is given by 1 d(H) = dim B ∈ M ( ) B¯ = BL, (BR) = (BR) ∀i, j R N C N i,ij j,ij

1 λj where Lij = f( ) and Ri,jk = f( ), f being the master function. λiλj λiλk Proof. The ﬁrst order deformation equations are: X ¯ HikHjk(Aik − Ajk) = 0 k n j ¯ nk With Hij = λi we have HijHjk = (λi/λj) , and so the defect is given by: ( ) λ nk λ nk X i X i d(H) = dimR A ∈ MN (R) Aik = Ajk ∀i, j λj λj k k t 1 ¯ Now, pick A ∈ MN (C) and set B = AH , so that A = N BH. First, we have: ¯ ¯ A ∈ MN (R) ⇐⇒ BH = BH 1 ⇐⇒ B¯ = BHH¯ ∗ N On the other hand, the matrix on the right is given by:

¯ ∗ X ¯ ¯ X −nk (HH )ij = HikHjk = (λiλj) = Lij k k HADAMARD MATRICES 77

¯ 1 Thus A ∈ MN (R) if and only the condition B = N BL in the statement is satisﬁed. 1 ¯ Regarding now the second condition on A, observe that with A = N BH we have: nk nk X λi 1 X λi Aik = Bis λj N λjλs k ks 1 X = B R N is s,ij s 1 = (BR) N i,ij

Thus the second condition on A reads (BR)i,ij = (BR)j,ij, and we are done. In view of the above results, a conjecture would be that the only isolated master Ha- damard matrices are the Fourier matrices Fp, with p prime. See [22].

Let us discuss now yet another interesting construction of complex Hadamard matrices, due to McNulty and Weigert [69]. The matrices constructed there generalize the Tao matrix T6, and usually have the interesting feature of being isolated. The construction in [69] uses the theory of MUB, as developed in [26], [48], but we will follow here a more direct approach, using basic Gauss sums, from [22]. The starting observation from [69] is as follows:

Theorem 5.14. Assuming that K ∈ MN (C) is Hadamard, so is the matrix 1 H = √ K (L∗R ) ia,jb Q ij i j ab √ √ provided that {L1,...,L√N } ⊂ QUQ and {R1,...,RN } ⊂ QUQ are such that each of √1 ∗ the matrices Q Li Rj ∈ QUQ, with i, j = 1,...,N, is Hadamard. Proof. The check of the unitarity is done as follows: 1 X < H ,H > = K (L∗R ) K¯ (L∗R ) ia kc Q ij i j ab kj k j cb jb X ¯ ∗ = KijKkj(Li Lk)ac j ∗ = Nδik(Li Lk)ac

= NQδikδac

The entries being in addition on the unit circle, we are done. As input for the above, we can use the following well-known Fourier construction: 78 TEO BANICA

q−1 Proposition 5.15. For q ≥ 3 prime, the matrices {Fq,DFq,...,D Fq}, where

2 3 6 10 q −1 10 6 3 D = diag 1, 1, w, w , w , w , . . . , w 8 , . . . , w , w , w , w

2πi/q √1 ∗ with w = e , are such that q Ei Ej is complex Hadamard, for any i 6= j. Proof. With 0, 1, . . . , q − 1 as indices, the formula of the above matrix D is:

0+1+...+(c−1) c(c−1) Dc = w = w 2 √ √1 ∗ Since we have q Ei Ej ∈ qUq, we just need to check that these matrices have entries belonging to T, for any i 6= j. With k = j − i, these entries are given by: 1 1 1 X √ (E∗E ) = √ (F ∗DkF ) = √ wc(b−a)Dk q i j ab q q q ab q c c Now observe that with s = b − a, we have the following formula: 2 X cs k X cs−ds c(c−1) ·k− d(d−1) ·k 2 2 w Dc = w w c cd X (c−d) c+d−1 ·k+s = w ( 2 ) cd X e 2d+e−1 ·k+s = w ( 2 ) de ! X e(e−1) ·k+es X edk = w 2 w e d X e(e−1) ·k+es = w 2 · qδe0 e = q

Thus the entries are on the unit circle, and we are done. We recall that the Legendre symbol is deﬁned as follows:  0 if s = 0 s  = 1 if ∃ α, s = α2 q −1 if 6∃ α, s = α2

Here, and in what follows, all the numbers are taken modulo q. We have: HADAMARD MATRICES 79

1 ∗ k c(c−1) √ 2 Proposition 5.16. The matrices Gk = q Fq D Fq, with D = diag(w ) being as above, and with k 6= 0 are circulant, their ﬁrst row vectors V k being given by

2 i ( i −1) k k/2 q −1 ·k − k k V = δ w 8 · w 2 i q q where δq = 1 if q = 1(4) and δq = i if q = 3(4), and with all inverses being taken in Zq.

Proof. This is a standard exercice on quadratic Gauss sums. First of all, the matrices Gk in the statement are indeed circulant, their ﬁrst vectors being given by:

k 1 X c(c−1) ·k+ic V = √ w 2 i q c Let us ﬁrst compute the square of this quantity. We have:

k 2 1 X c(c−1) + d(d−1) k+i(c+d) (V ) = w[ 2 2 ] i q cd The point now is that the sum S on the right, which has q2 terms, decomposes as follows, where x is a certain exponent, depending on q, i, k: ( (q − 1)(1 + w + ... + wq−1) + qwx if q = 1(4) S = (q + 1)(1 + w + ... + wq−1) − qwx if q = 3(4)

We conclude that we have a formula as follows, where δq ∈ {1, i} is as in the statement, 2 2 2 so that δq ∈ {1, −1} is given by δq = 1 if q = 1(4) and δq = −1 if q = 3(4): k 2 2 x (Vi ) = δq w In order to compute now the exponent x, we must go back to the above calculation of the sum S. We succesively have: q2−1 – First of all, at k = 1, i = 0 we have x = 4 . q2−1 – By translation we obtain x = 4 − i(i − 1), at k = 1 and any i. k q2−1 i i – By replacing w → w we obtain x = 4 · k − k ( k − 1), at any k 6= 0 and any i. Summarizing, we have computed the square of the quantity that we are interested in, the formula being as follows, with δq being as in the statement:

q2−1 i i k 2 2 4 ·k − k ( k −1) (Vi ) = δq · w · w By extracting now the square root, we obtain a formula as follows:

2 i ( i −1) q −1 k k k 8 ·k − 2 Vi = ±δq · w · w The computation of the missing sign is non-trivial, but by using the theory of quadratic Gauss sums, and more speciﬁcally a result of Gauss, computing precisely this kind of sign, k/2 we conclude that we have indeed a Legendre symbol, ± = q , as claimed. 80 TEO BANICA

Let us combine now all the above results. We obtain the following statement: Theorem 5.17. Let q ≥ 3 be prime, consider two subsets S,T ⊂ {0, 1, . . . , q−1} satisfying |S| = |T | and S ∩ T = ∅, and write S = {s1, . . . , sN } and T = {t1, . . . , tN }. Then

tj −si Hia,jb = KijVb−a where V is as above, is Hadamard, provided that K ∈ MN (C) is. Proof. This follows indeed by using the general construction in Theorem 5.14 above, with input coming from Proposition 5.15 and Proposition 5.16. As explained in [69], the above construction covers many interesting examples of Ha- damard matrices, known from [88], [89] to be isolated, such as the Tao matrix: 1 1 1 1 1 1  2 2 1 1 w w w w   2 2  1 w 1 w w w  T6 =  2 2 1 w w 1 w w  1 w2 w2 w 1 w  1 w2 w w2 w 1 In general, in order to ﬁnd isolated matrices, the idea from [69] is that of starting with an isolated matrix, and then use suitable sets S,T . The defect computations are, however, quite diﬃcult. As a concrete statement, however, we have the following conjecture: Conjecture 5.18. The complex Hadamard matrix constructed in Theorem 5.17 is isolated, provided that: (1) K is an isolated Fourier matrix, of prime order. (2) S,T consist of consecutive odd numbers, and consecutive even numbers.

This statement is supported by the isolation result for T6, and by several computer simulations in [69]. For further details on all this, we refer to [69], and to [22].

As a ﬁnal topic now, we would like to discuss an extension of a part of the above results, to the case of the partial Hadamard matrices. The extension, done in [22], is quite straightforward, but there are however a number of subtleties appearing. First of all, we can talk about deformations of PHM, as follows:

Deﬁnition 5.19. Let H ∈ XM,N be a partial complex Hadamard matrix. (1) A deformation of H is a smooth function f : T1 → (XM,N )H . Aij (2) The deformation is called “aﬃne” if fij(q) = Hijq , with A ∈ MM×N (R). ai+bj M N (3) We call “trivial” the deformations fij(q) = Hijq , with a ∈ R , b ∈ R . √ We have XM,N = MM×N (T)∩ NUM,N , where UM,N ⊂ MM×N (C) is the set of matrices having all rows of norm 1, and pairwise orthogonal. This remark leads us to: HADAMARD MATRICES 81

Definition 5.20. Associated to a point H ∈ XM,N are: √ (1) The enveloping tangent space: TeH XM,N = TH MM×N (T) ∩ TH NUM,N . (2) The tangent cone TH XM,N : the set of tangent vectors to the deformations of H. ◦ (3) The affine tangent cone TH XM,N : same as above, using affine deformations only. × (4) The trivial tangent cone TH XM,N : as above, using trivial deformations only.

◦ Observe that TeH XM,N ,TH XM,N are real vector spaces, and that TH XM,N ,TH XM,N are two-sided cones, λ ∈ R,A ∈ T =⇒ λA ∈ T . Also, we have inclusions as follows: × ◦ TH XM,N ⊂ TH XM,N ⊂ TH XM,N ⊂ TeH XM,N

Since TeH XM,N is a real vector space, of particular interest is the computation of its dimension d(H) = dim(TeH XM,N ), called defect of H. We have: √ Theorem 5.21. Let H ∈ XM,N , and pick K ∈ NUN extending H. P ¯ (1) TeH XM,N '{A ∈ MM×N (R)| k HikHjk(Aik − Ajk) = 0, ∀i, j}. ∗ ¯ (2) TeH XM,N '{E = (XY ) ∈ MM×N (C)|X = X , (EK)ijHij ∈ R, ∀i, j}. P ¯ ¯ The correspondence A → E is given by Eij = k HikKjkAik, Aij = (EK)ijHij. Proof. The proofs here go as in the square case, as follows: (1) In the square case this was done in the proof of Theorem 4.4 above, and the extension of the computations there to the rectangular case is straightforward. ∗ (2) Let us set indeed Rij = AijHij and E = RK . The correspondence A → R → E is then bijective, and we have the following formula: X ¯ Eij = HikKjkAik k ¯ With these changes, the system of equations in (1) becomes Eij = Eji for any i, j with j ≤ M. But this shows that we must have E = (XY ) with X = X∗, and the condition ¯ Aij ∈ R corresponds to the condition (EK)ijHij ∈ R, as claimed. As an illustration, in the real case we obtain the following result:

Theorem 5.22. For an Hadamard matrix H ∈ MM×N (±1) we have

symm TeH XM,N ' MM (R) ⊕ MM×(N−M)(R) and so the defect is given by d(H) = N(N + 1)/2 + M(N − M) independently of the precise value of H. 82 TEO BANICA √ Proof. We use Theorem 5.21 (2). Since H is now real we can pick K ∈ NUN extending it to be real too, and with nonzero entries, so the last condition appearing there, namely ¯ (EK)ijHij ∈ R, simply tells us that E must be real. Thus we have: ∗ TeH XM,N '{E = (XY ) ∈ MM×N (R)|X = X } But this is the formula in the statement, and we are done.

A matrix H ∈ XM,N cannot be isolated, simply because the space of its Hadamard MN equivalents provides a copy T ⊂ XM,N , passing through H. However, if we restrict the attention to the matrices which are dephased, the notion of isolation makes sense:

Proposition 5.23. Let d(H) = dim(TeH XM,N ). (1) This number, called undephased defect of H, satisﬁes d(H) ≥ M + N − 1. (2) If d(H) = M+N−1 then H is isolated inside the dephased quotient XM,N → ZM,N . Proof. Once again, the known results in the square case extend: × (1) We have indeed dim(TH XM,N ) = M + N − 1, and since the tangent vectors to these trivial deformations belong to TeH XM,N , this gives the result. × (2) Since d(H) = M + N − 1, the inclusions TH XM,N ⊂ TH XM,N ⊂ TeH XM,N must be × equalities, and from TH XM,N = TH XM,N we obtain the result. Finally, still at the theoretical level, we have the following conjecture:

Conjecture 5.24. An isolated matrix H ∈ ZM,N must have minimal defect, namely d(H) = M + N − 1.

In other words, the conjecture is that if H ∈ ZM,N has only trivial ﬁrst order deformations, then it has only trivial deformations at any order, including at ∞. In the square matrix case this statement comes with solid evidence, all known examples of complex Hadamard matrices H ∈ XN having non-minimal defect being known to admit one-parameter deformations. For more on this subject, see [88], [89].

Let us discuss now some examples of isolated partial Hadamard matrices, and provide some evidence for Conjecture 5.24. We are interested in the following matrices:

Deﬁnition 5.25. The truncated Fourier matrix FS,G, with G being a ﬁnite abelian group, and with S ⊂ G being a subset, is constructed as follows: ij 2πi/N (1) Given N ∈ N, we set FN = (w )ij, where w = e .

(2) Assuming G = ZN1 × ... × ZNs , we set FG = FN1 ⊗ ... ⊗ FNs . (3) We let FS,G be the submatrix of FG having S ⊂ G as row index set.

Observe that FN is the Fourier matrix of the cyclic group ZN . More generally, FG is the Fourier matrix of the ﬁnite abelian group G. Observe also that FG,G = FG. We can compute the defect of FS,G by using Theorem 5.21, and we obtain: HADAMARD MATRICES 83

Theorem 5.26. For a truncated Fourier matrix F = FS,G we have the formula n o t ¯ TeF XM,N = A ∈ MM×N (R) P = AF satisﬁes Pij = Pi+j,j = Pi,−j, ∀i, j where M = |S|,N = |G|, and with all the indices being regarded as group elements. Proof. We use Theorem 5.21 (1). The defect equations there are as follows: X ¯ FikFjk(Aik − Ajk) = 0 k ¯ t Since for F = FS,G we have FikFjk = (F )k,i−j, we obtain: n o t t TeF XM,N = A ∈ MM×N (R) (AF )i,i−j = (AF )j,i−j, ∀i, j

Now observe that for an arbitrary matrix P ∈ MM (C), we have:

Pi,i−j = Pj,i−j, ∀i, j ⇐⇒ Pi+j,i = Pji, ∀i, j

⇐⇒ Pi+j,j = Pij, ∀i, j We therefore conclude that we have the following equality: n o t TeF XM,N = A ∈ MM×N (R) P = AF satisﬁes Pij = Pi+j,j, ∀i, j t Now observe that with A ∈ MM×N (R) and P = AF ∈ MM (C) as above, we have: ¯ X ∗ Pij = Aik(F )kj k X t = Aik(F )k,−j k = Pi,−j Thus, we obtain the formula in the statement, and we are done. Let us try to ﬁnd some explicit examples of isolated matrices, of truncated Fourier type. For this purpose, we can use the following improved version of Theorem 5.26:

Theorem 5.27. The defect of F = FS,G is the number dim(K) + dim(I), where n o t K = A ∈ MM×N (R) AF = 0 n o t I = P ∈ LM ∃A ∈ MM×N (R),P = AF where LM is the following linear space ¯ LM = P ∈ MM (C) Pij = Pi+j,j = Pi,−j, ∀i, j with all the indices belonging by deﬁnition to the group G. 84 TEO BANICA

Proof. We use the general formula in Theorem 5.26. With the notations there, and with the linear space LM being as above, we have a linear map as follows: t Φ: TeF XM,N → LM , Φ(A) = AF By using this map, we obtain the following equality:

dim(TeF XM,N ) = dim(ker Φ) + dim(Im Φ) Now since the spaces on the right are precisely those in the statement, ker Φ = K and Im Φ = I, by applying Theorem 5.26 we obtain the result. In order to look now for isolated matrices, the ﬁrst remark is that since a deformation of FG will produce a deformation of FS,G too, we must restrict the attention to the case where G = Zp, with p prime. And here, we have the following conjecture:

Conjecture 5.28. There exists a constant ε > 0 such that FS,p is isolated, for any p prime, once S ⊂ Zp satisfies |S| ≥ (1 − ε)p. In principle this conjecture can be approached by using the formula in Theorem 5.27, and we have for instance evidence towards the fact that Fp−1,p should be always isolated, that Fp−2,p should be isolated too, provided that p is big enough, and so on. However, finding a number ε > 0 as above looks like a quite difficult question. See [22]. HADAMARD MATRICES 85

6. Circulant matrices We discuss in this section yet another type of special complex Hadamard matrices, namely the circulant ones. There has been a lot of work here, starting with the Circulant Hadamard Conjecture (CHC) in the real case, and with many results in the complex case as well. We will present here the main techniques in dealing with such matrices. It is convenient to introduce the circulant matrices as follows:

Deﬁnition 6.1. A complex matrix H ∈ MN (C) is called circulant when we have

Hij = γj−i for some γ ∈ CN , with the matrix indices i, j ∈ {0, 1,...,N − 1} taken modulo N.

Here the index convention is quite standard, as for the Fourier matrices FN , and with this coming from Fourier analysis considerations, that we will get into later on. Here is a basic, and very fundamental example of a circulant Hadamard matrix, which in addition has real entries, and is symmetric: −1 1 1 1   1 −1 1 1  K4 =    1 1 −1 1  1 1 1 −1 According to the CHC, explained in section 1, this matrix is, up to equivalence, the only circulant Hadamard matrix H ∈ MN (±1), regardless of the value of N ∈ N. Our ﬁrst purpose will be that of showing that the CHC dissapears in the complex case, where we have examples at any N ∈ N. As a ﬁrst result here, we have: Proposition 6.2. The following are circulant and symmetric Hadamard matrices, −1 ν 1 ν  w 1 1 i 1 ν −1 ν 1 F 0 = ,F 0 = 1 w 1 ,F 00 =   2 1 i 3   4  1 ν −1 ν  1 1 w   ν 1 ν −1

2πi/3 πi/4 where w = e , ν = e , equivalent to the Fourier matrices F2,F3,F4. Proof. The orthogonality between rows being clear, we have here complex Hadamard 0 matrices. The fact that we have an equivalence F2 ∼ F2 follows from: 1 1 i i i 1 ∼ ∼ 1 −1 1 −1 1 i 0 At N = 3 now, the equivalence F3 ∼ F3 can be constructed as follows: 1 1 1  1 1 w w 1 1 1 w w2 ∼ 1 w 1 ∼ 1 w 1 1 w2 w w 1 1 1 1 w 86 TEO BANICA

00 As for the case N = 4, here the equivalence F4 ∼ F4 can be constructed as follows, 2πki/s where we use the logarithmic notation [k]s = e , with respect to s = 8: 0 0 0 0 0 1 4 1 4 1 0 1 0 2 4 6 1 4 1 0 1 4 1 0   ∼   ∼   0 4 0 4 4 1 0 1 0 1 4 1 0 6 4 2 1 0 1 4 1 0 1 4 8 8 8 00 0 We will explain later the reasons for denoting this matrix F4 , instead of F4.

Getting back now to the real circulant matrix K4, this is equivalent to the Fourier matrix FG = F2 ⊗ F2 of the Klein group G = Z2 × Z2, as shown by: −1 1 1 1   1 1 1 −1 1 1 1 1   1 −1 1 1   1 −1 1 1  1 −1 1 −1   ∼   ∼    1 1 −1 1   1 1 −1 1  1 1 −1 −1 1 1 1 −1 −1 1 1 1 1 −1 −1 1

In fact, we have the following construction of circulant and symmetric Hadamard matrices at N = 4, which involves an extra parameter q ∈ T: Proposition 6.3. The following circulant and symmetric matrix is Hadamard,

−1 q 1 q  q  q −1 q 1  K =   4  1 q −1 q  q 1 q −1

πi/4 00 for any q ∈ T. At q = 1, e recover respectively the matrices K4,F4 .

Proof. The rows of the above matrix are pairwise orthogonal for any q ∈ C, and so at q ∈ T we obtain a complex Hadamard matrix. The last assertion is clear. As a ﬁrst conclusion, coming from the above considerations, we have:

Theorem 6.4. The complex Hadamard matrices of order N = 2, 3, 4, 5, namely

p F2,F3,F4 ,F5 can be put, up to equivalence, in circulant and symmetric form.

Proof. As explained in section 2 above, according to the result of Haagerup [51], the Hadamard matrices at N = 2, 3, 4, 5 are, up to equivalence, those in the statement. At N = 2, 3 the problem is solved by Proposition 6.2 above. HADAMARD MATRICES 87

q s −2 At N = 4 now, our claim is that we have K4 ∼ F4 , with s = q . Indeed, by multiplying q the rows of K4 , and then the columns, by suitable scalars, we have: −1 q 1 q  1 −q −1 −q 1 1 1 1  q  q −1 q 1  1 −q¯ 1q ¯  1 s −1 −s K =   ∼   ∼   4  1 q −1 q  1 q −1 q  1 −1 1 −1 q 1 q −1 1q ¯ 1 −q¯ 1 −s −1 s

s On the other hand, by permuting the second and third rows of F4 , we obtain: 1 1 1 1  1 1 1 1  s 1 −1 1 −1 1 s −1 −s F =   ∼   4 1 s −1 −s 1 −1 1 −1 1 −s −1 s 1 −s −1 s Thus these matrices are equivalent, and the result follows from Proposition 6.3. At N = 5, the matrix that we are looking for is as follows, with w = e2πi/5: w2 1 w4 w4 1   1 w2 1 w4 w4 0  4 2 4 F5 = w 1 w 1 w   4 4 2  w w 1 w 1  1 w4 w4 1 w2 It is indeed clear that this matrix is circulant, symmetric, and complex Hadamard, and 0 the fact that we have F5 ∼ F5 follows either directly, or by using [51].

Let us prove now that any Fourier matrix FN can be put in circulant and symmetric form. We use Bj¨orck’s cyclic root formalism [30], which is as follows:

Theorem 6.5. Assume that H ∈ MN (T) is circulant, Hij = γj−i. Then H is Hadamard if and only if the vector (z0, z1, . . . , zN−1) given by zi = γi/γi−1 satisﬁes:

z0 + z1 + ... + zN−1 = 0

z0z1 + z1z2 + ... + zN−1z0 = 0 ...

z0z1 . . . zN−2 + ... + zN−1z0 . . . zN−3 = 0

z0z1 . . . zN−1 = 1

If so is the case, we say that z = (z0, . . . , zN−1) is a cyclic N-root.

Proof. This follows from a direct computation, the idea being that, with Hij = γj−i as above, the orthogonality conditions between the rows are best written in terms of the variables zi = γi/γi−1, and correspond to the equations in the statement. See [30]. 88 TEO BANICA

Observe that, up to a global multiplication by a scalar w ∈ T, the ﬁrst row vector γ = (γ0, . . . , γN−1) of the matrix H ∈ MN (T) constructed above is as follows:

γ = (z0, z0z1, z0z1z2, ...... , z0z1 . . . zN−1) Now back to the Fourier matrices, we have the following result:

Theorem 6.6. Given N ∈ N, set ν = eπi/N and q = νN−1, w = ν2. Then (q, qw, qw2, . . . , qwN−1) 0 is a cyclic N-root, and the corresponding complex Hadamard matrix FN is circulant and symmetric, and equivalent to the Fourier matrix FN . Proof. Given q, w ∈ T, let us ﬁnd out when (q, qw, qw2, . . . , qwN−1) is a cyclic root: (1) In order for the = 0 equations in Theorem 6.5 to be satisﬁed, the value of q is irrelevant, and w must be a primitive N-root of unity. (2) As for the = 1 equation in Theorem 6.5, this states in our case that we must have N N(N−1) N N−1 q w 2 = 1, and so that we must have q = (−1) . We conclude that with the values of q, w ∈ T in the statement, we have indeed a cyclic N-root. Now construct Hij = γj−i as in Theorem 6.5. We have:

k+1 k(k+1) −k+1 k(k−1) γk = γ−k, ∀k ⇐⇒ q w 2 = q w 2 , ∀k ⇐⇒ q2kwk = 1, ∀k ⇐⇒ q2 = w−1 But this latter condition holds indeed, because we have q2 = ν2N−2 = ν−2 = w−1. We conclude that our circulant matrix H is symmetric as well, as claimed. It remains to construct an equivalence H ∼ FN . In order to do this, observe that, due to our conventions q = νN−1, w = ν2, the ﬁrst row vector of H is given by:

k+1 k(k+1) γk = q w 2 = ν(N−1)(k+1)νk(k+1) = ν(N+k−1)(k+1) Thus, the entries of H are given by the following formula:

H−i,j = H0,i+j = ν(N+i+j−1)(i+j+1) = νi2+j2+2ij+Ni+Nj+N−1 = νN−1 · νi2+Ni · νj2+Nj · ν2ij

With this formula in hand, we can now ﬁnish. Indeed, the matrix H = (Hij) is 0 0 equivalent to the matrix H = (H−i,j). Now regarding H , observe that in the above formula, the factors νN−1, νi2+Ni, νj2+Nj correspond respectively to a global multiplication HADAMARD MATRICES 89 by a scalar, and to row and column multiplications by scalars. Thus H0 is equivalent to the matrix H00 obtained by deleting these factors. 00 2ij πi/N But this latter matrix, given by Hij = ν with ν = e , is precisely the Fourier matrix FN , and we are done. As an illustration, let us work out the cases N = 2, 3, 4, 5. We have here:

0 Proposition 6.7. The matrices FN are as follows: 0 0 (1) At N = 2, 3 we obtain the old matrices F2,F3. (2) At N = 4 we obtain the following matrix, with ν = eπi/4: ν3 1 ν7 1  3 7 0  1 ν 1 ν  F =  7 3  4 ν 1 ν 1  1 ν7 1 ν3

0 (3) At N = 5 we obtain the old matrix F5. Proof. With notations from Theorem 6.6, the proof goes as follows: (1) At N = 2 we have ν = i, q = i, w = −1, so the cyclic root is (i, −i), the first row 0 πi/3 vector is (i, 1), and we obtain indeed the old matrix F2. At N = 3 we have ν = e and q = w = ν2 = e2πi/3, the cyclic root is (w, w2, 1), the first row vector is (w, 1, 1), and we 0 obtain indeed the old matrix F3. (2) At N = 4 we have ν = eπi/4 and q = ν3, w = ν2, the cyclic root is (ν3, ν5, ν7, ν), the first row vector is (ν3, 1, ν7, 1), and we obtain the matrix in the statement. (3) At N = 5 we have ν = eπi/5 and q = ν4 = w2, with w = ν2 = e2πi/5, the cyclic root is therefore (w2, w3, w4, 1, w), the first row vector is (w2, 1, w4, w4, 1), and we obtain 0 in this way the old matrix F5, as claimed. 0 00 Regarding the above matrix F4, observe that this is equivalent to the matrix F4 from 0 00 Proposition 6.2, with the equivalence F4 ∼ F4 being obtained by multiplying everything by ν = eπi/4. While both these matrices are circulant and symmetric, and of course equiv- 0 alent to F4, one of them, namely F4, is “better” than the other, because the corresponding 0 00 cyclic root comes from a progression. This is the reason for our notations F4,F4 .

Let us discuss now the case of the generalized Fourier matrices FG. In this context, the assumption of being circulant is somewhat unnatural, because this comes from a ZN symmetry, and the underlying group is no longer ZN . It is possible to ﬁx this issue by talking about G-patterned Hadamard matrices, with G being no longer cyclic, but for our purposes here, best is to formulate the result in a weaker form, as follows:

Theorem 6.8. The generalized Fourier matrices FG, associated to the ﬁnite abelian groups G, can be put in symmetric and bistochastic form. 90 TEO BANICA

Proof. We know from Theorem 6.6 that any usual Fourier matrix FN can be put in circulant and symmetric form. Since circulant implies bistochastic, in the sense that the sums on all rows and all columns must be equal, the result holds for FN .

In general now, if we decompose G = ZN1 × ... × ZNk , we have:

FG = FN1 ⊗ ... ⊗ FNk Now since the property of being circulant is stable under taking tensor products, and so is the property of being bistochastic, we therefore obtain the result. We have as well the following alternative generalization of Theorem 6.6, coming from Backelin’s work in [3], and remaining in the circulant and symmetric setting: Theorem 6.9. Let M|N, and set w = e2πi/N . We have a cyclic root as follows, N−1 N−1 (q1, . . . , qM , q1w, . . . , qM w, ...... , q1w , . . . , qM w ) | {z } | {z } | {z } M M M N M(N−1) provided that q1, . . . , qM ∈ T satisfy (q1 . . . qM ) = (−1) . Moreover, assuming

q1q2 = 1 , q3qM = q4qM−1 = ... = w N M(N−1) which imply (q1 . . . qM ) = (−1) , the Hadamard matrix is symmetric. Proof. Let us ﬁrst check the = 0 equations for a cyclic root. Given arbitrary numbers q1, . . . , qM ∈ T, if we denote by (zi) the vector in the statement, we have: X q1 . . . qK + q2 . . . qK+1 + ...... + qM−K+1 . . . qM zi+1 . . . zi+K = K−1 +qM−K+2 . . . qM q1w + ...... + qM q1 . . . qK−1w i ×(1 + wK + w2K + ... + w(N−1)K ) Now since the sum on the right vanishes, the = 0 conditions are satisﬁed. Regarding now the = 1 condition, the total product of the numbers zi is given by:

Y N 2 N−1 M N MN(N−1) zi = (q1 . . . qM ) (1 · w · w . . . w ) = (q1 . . . qM ) w 2 i By using w = e2πi/N we obtain that the coeﬃcient on the right is:

MN(N−1) 2πi · MN(N−1) πiM(N−1) M(N−1) w 2 = e N 2 = e = (−1) N M(N−1) Thus, if (q1 . . . qM ) = (−1) , we obtain a cyclic root, as stated. See [3], [49]. The corresponding ﬁrst row vector can be written as follows:    wM−1 w2 w  V = q1, q1q2, . . . , q1 . . . qM ,...... , ,..., , , 1 | {z } q2 . . . qM qM−1qM qM  M | {z } M HADAMARD MATRICES 91

Thus, the corresponding circulant complex Hadamard matrix is as follows:   q1 q1q2 q1q2q3 q1q2q3q4 q1q2q3q4q5 ...  1 q1 q1q2 q1q2q3 q1q2q3q4 ...  w  1 q1 q1q2 q1q2q3 ...  qM  H =  w2 w   1 q1 q1q2 ...  qM−1qM qM   w3 w2 w  1 q1 ...  qM−2qM−1qM qM−1qM qM  ...... We are therefore led to the symmetry conditions in the statement, and we are done. Observe that Theorem 6.9 generalizes both Proposition 6.3, and the construction in Theorem 6.6. Thus, we have here a full generalization of Theorem 6.4. Of course, things do not stop here, and the problem of unifying Theorem 6.8 and Theorem 6.9 remains. As a conclusion to what we have so far, there is deﬁnitely no analogue of the CHC in the general complex setting, due to the fact that FN can be put in circulant form, and this latter result gives rise to a number of interesting generalizations and questions. However, still in relation with the CHC, but at a more technical level, the problem of investigating the existence of the circulant Butson matrices appears. The ﬁrst result in this direction, due to Turyn [93], is as follows: Proposition 6.10. The size of a circulant Hadamard matrix

H ∈ MN (±1) must be of the form N = 4n2, with n ∈ N. Proof. Let a, b ∈ N with a + b = N be the number of 1, −1 entries in the ﬁrst row of H. If we denote by H0,...,HN−1 the rows of H, then by summing over columns we get: N−1 X < H0,Hi > = a(a − b) + b(b − a) i=0 = (a − b)2

On the other hand, the quantity on the left is < H0,H0 >= N. Thus N is a square, 2 and together with the fact that N ∈ 2N, this gives N = 4n , with n ∈ N. Also found by Turyn in [93] is the fact that the above number n ∈ N must be odd, and not a prime power. In the general Butson matrix setting now, we have: 2πi/l Proposition 6.11. Assume that H ∈ HN (l) is circulant, let w = e . If a0, . . . , al−1 ∈ P l−1 N with ai = N are the number of 1, w, . . . , w entries in the ﬁrst row of H, then: X k w aiai+k = N ik P This condition, with ai = N, will be called “Turyn obstruction” on (N, l). 92 TEO BANICA

Proof. Indeed, by summing over the columns of H, we obtain: X X i j < H0,Hi > = < w , w > aiaj i ij X i−j = w aiaj ij

Now since the left term is < H0,H0 >= N, this gives the result. We can deduce from this a number of concrete obstructions, as follows: P 2 Theorem 6.12. When l is prime, the Turyn obstruction is i(ai − ai+k) = 2N for any k 6= 0. Also, for small values of l, the Turyn obstruction is as follows: 2 (1) At l = 2 the condition is (a0 − a1) = N. 2 2 2 (2) At l = 3 the condition is (a0 − a1) + (a1 − a2) + (a2 − a3) = 2N. 2 2 (3) At l = 4 the condition is (a0 − a2) + (a1 − a3) = N. P 2 P 2 (4) At l = 5 the condition is i(ai − ai+1) = i(ai − ai+2) = 2N. Proof. We use the fact, from Proposition 3.3 above, that when l is prime, the vanishing sums of l-roots of unity are exactly the sums of the following type, with c ∈ N: S = c + cw + ... + cwl−1 Thus the Turyn obstruction is equivalent to the following equations, one for each k 6= 0: X 2 X ai − aiai+k = N i i Now by forming squares, this gives the equations in the statement. Regarding now the l = 2, 3, 4, 5 assertions, these follow from the ﬁrst assertion when l is prime, l = 2, 3, 5. Also, at l = 4 we have w = i, so the Turyn obstruction reads: 2 2 2 2 X X (a0 + a1 + a2 + a3) + i aiai+1 − 2(a0a2 + a1a3) − i aiai+1 = N Thus the imaginary terms cancel, and we obtain the formula in the statement. The above results are of course just some basic, elementary observations on the subject, and the massive amount of work on the CHC has a number of interesting Butson matrix extensions. For some more advanced theory on all this, we refer to [18], [39]. Let us go back now to the pure complex case, and discuss Fourier analytic aspects. From a traditional linear algebra viewpoint, the circulant matrices are best understood as being the matrices which are Fourier-diagonal,√ and we will exploit this here. ij 2πi/N Let us ﬁx N ∈ N, and denote√ by F = (w )/ N with w = e the rescaled Fourier matrix. Observe that FN = NF is the usual Fourier Hadamard matrix. N Given a vector q ∈ C , we denote by Q ∈ MN (C) the diagonal matrix having q as vector of diagonal entries. That is, Qii = qi, and Qij = 0 for i 6= j. With these conventions, the above-mentioned linear algebra result is as follows: HADAMARD MATRICES 93

Theorem 6.13. For a complex matrix H ∈ MN (C), the following are equivalent: N (1) H is circulant, Hij = ξj−i for some ξ ∈ C . (2) H is Fourier-diagonal, H = F QF ∗ with Q diagonal. √ In addition, the ﬁrst row vector of F QF ∗ is given by ξ = F q/ N.

∗ Proof. If Hij = ξj−i is circulant then Q = F HF is diagonal, given by: 1 X X Q = wjl−ikξ = δ wjrξ ij N l−k ij r kl r Also, if Q = diag(q) is diagonal then H = F QF ∗ is circulant, given by: X 1 X H = F Q F¯ = w(i−j)kq ij ik kk jk N k k k √ Observe that this latter formula proves as well the last assertion, ξ = F q/ N. In relation now with the orthogonal and unitary matrices, we have: Proposition 6.14. The various sets of circulant matrices are as follows: circ ∗ N (1) MN (C) = {F QF |q ∈ C }. circ ∗ N (2) UN = {F QF |q ∈ T }. (3) Ocirc = {F QF ∗|q ∈ N , q¯ = q , ∀i}. N T i −i √ In addition, the ﬁrst row vector of F QF ∗ is given by ξ = F q/ N. Proof. All this follows from Theorem 6.13, as follows: (1) This assertion, along with the last one, is Theorem 6.13 itself. (2) This is clear from (1), because the eigenvalues must be on the unit circle T. N (3) Observe ﬁrst that for q ∈ C we have F q = F q˜, withq ˜i =q ¯−i, and so ξ = F q is real if and only ifq ¯i = q−i for any i. Together with (2), this gives the result.

Observe that in (3), the equations for the parameter space are q0 =q ¯0,q ¯1 = qn−1, q¯2 = qn−2, and so on until [N/2] + 1. Thus, with the convention Z∞ = T we have: ( × (N−1)/2 (N odd) Ocirc ' Z2 Z∞ N 2 (N−2)/2 Z2 × Z∞ (N even) In terms of circulant Hadamard matrices, we have the following statement: Theorem 6.15. The sets of complex and real circulant Hadamard matrices are: √ circ ∗ N XN = { NF QF |q ∈ T } ∩ MN (T) √ circ ∗ N YN = { NF QF |q ∈ T , q¯i = q−i} ∩ MN (±1) In addition, the sets of q parameters are invariant under cyclic permutations, and also under mutiplying by numbers in T, respectively under multiplying by −1. 94 TEO BANICA

Proof. All the assertions are indeed clear from Proposition 6.14 above. The above statement is of course something quite theoretical in the real case, where circ the CHC states that we should have YN = ∅, at any N 6= 4. However, in the complex case all this is useful, and complementary to Bj¨orck’s cyclic root formalism.

Let us discuss now a number of geometric and analytic aspects. First, we have the following deep counting result, due to Haagerup [52]: Theorem 6.16. When N is prime, the number of circulant N × N complex Hadamard 2N−2 matrices, counted with certain multiplicities, is exactly N−1 . Proof. This is something advanced, using a variety of techiques from Fourier analysis, number theory, complex analysis and algebraic geometry. The idea in [52] is that, when N is prime, Björck’s cyclic root formalism can be further manipulated, by using Fourier transforms, and we are eventually led to a simpler system of equations. This simplified system can be shown then to have a finite number of solutions, the key ingredient here being a well-known theorem of Chebotarev, which states that when N is prime, all the minors of the Fourier matrix FN are nonzero. With this finiteness result in hand, the precise count can be done as well, by using various techniques from classical algebraic geometry. See [52]. When N is not prime, the situation is considerably more complicated, with some values leading to finitely many solutions, and with other values leading to an infinite number of solutions, and with many other new phenomena appearing. See [30], [31], [32], [52]. Our belief is that useful here would be an adaptation of the notion of defect, to the circulant case, in the context of the manifolds from Proposition 6.14 and Theorem 6.15. There are some preliminary differential geometry computations to be done here.

We would like to√ discuss now an alternative take on these questions, based on the estimate ||U||1 ≤ N N from Theorem 1.18. This shows that the real Hadamard matrices are the rescaled versions of the maximizers of the 1-norm on ON , and the same proof shows that the complex Hadamard matrices are the rescaled versions of the maximizers of the 1-norm on UN . Following [18], we will apply this philosophy to the circulant case. We will need in fact more general p-norms as well, so let us start with the following result, in the complex case, which is the most general one on the subject:

Proposition 6.17. If ψ : [0, ∞) → R is strictly concave/convex, the quantity

X 2 F (U) = ψ(|Uij| ) ij over UN is maximized/minimized precisely by the rescaled Hadamard matrices. HADAMARD MATRICES 95

Proof. We recall that Jensen’s inequality states that for ψ convex we have: x + ... + x ψ(x ) + ... + ψ(x ) ψ 1 n ≤ 1 n n n For ψ concave the reverse inequality holds. Also, the equality case holds either when ψ is linear, or when the numbers x1, . . . , xn are all equal. 2 2 In our case, with n = N and with {x1, . . . , xn} = {|Uij| |i, j = 1,...,N}, we obtain that for any convex function ψ, the following holds: 1 F (U) ψ ≤ N N 2 Thus we have F (U) ≥ N 2ψ(1/N), and by assuming as in the statement that ψ is 2 strictly convex,√ the equality case holds precisely when the numbers |Uij| are all equal, so when H = NU is Hadamard. The proof for concave functions is similar. Of particular interest for our considerations are the power functions ψ(x) = xp/2, which are concave at p ∈ [1, 2), and convex at p ∈ (2, ∞). These lead to: √ Theorem 6.18. The rescaled versions U = H/ N of the complex Hadamard matrices H ∈ MN (C) can be characterized as being: (1) The maximizers of the p-norm on UN , at any p ∈ [1, 2). (2) The minimizers of the p-norm on UN , at any p ∈ (2, ∞].

Proof. Consider indeed the p-norm on UN , which at p ∈ [1, ∞) is given by: !1/p X p ||U||p = |Uij| ij By the above discussion, involving the functions ψ(x) = xp/2, Proposition 6.17 applies and gives the results at p ∈ [1, ∞), the precise estimates being as follows:  ≤ N 2/p−1/2 if p < 2  1/2 ||U||p = = N if p = 2 ≥ N 2/p−1/2 if p > 2 As for the case p = ∞, this follows with p → ∞, or directly via Cauchy-Schwarz. As explained in [18], the most adapted exponent for the circulant case is p = 4, due to a number of simpliﬁcations which appear in the Fourier manipulations. So, before discussing this, let us record the p = 4 particular case of Theorem 6.18:

Proposition 6.19. Given a matrix U ∈ UN we have

||U||4 ≥ 1 √ with equality precisely when H = U/ N is Hadamard. 96 TEO BANICA

Proof. This follows from Theorem 6.18, or directly from Cauchy-Schwarz, as follows: !2 X 1 X ||U||4 = |U |4 ≥ |U |2 = 1 4 ij N 2 ij ij ij √ Thus we have ||U||4 ≥ 1, with equality if and only if H = NU is Hadamard. In the circulant case now, and in Fourier formulation, the estimate is as follows:

N Theorem 6.20. Given a vector q ∈ T , written q = (q0, . . . , qN−1) consider the following quantity, with all the indices being taken modulo N:

X qiqk Φ = qjql i+k=j+l √ Then Φ is real, and we have Φ ≥ N 2, with equality if and only if Nq is the eigenvalue vector of a circulant Hadamard matrix H ∈ MN (C). Proof. By conjugating the formula of Φ we see that this quantity is indeed real. In fact, Φ appears by deﬁnition as a sum of N 3 terms, consisting of N(2N − 1) values of 1 and of N(N − 1)2 other complex numbers of modulus 1, coming in pairs (a, a¯). Regarding now the second assertion, by using the√ various identiﬁcations in Theorem 6.13 and Proposition 6.14, and the formula ξ = F q/ N there, we have:

4 X 4 ||U||4 = N |ξs| s 1 X X = | wsiq |4 N 3 i s i 1 X X X X X = wsiq w−sjq¯ wskq w−slq¯ N 3 i j k l s i j k l

1 X X (i−j+k−l)s qiqk = 3 w N qjql s ijkl

1 X qiqk = 2 N qjql i+k=j+l Thus Proposition 6.19 gives the following estimate:

2 4 2 Φ = N ||U||4 ≥ N Moreover, we have equality precisely in the Hadamard matrix case, as claimed. We have the following more direct explanation of the above result: HADAMARD MATRICES 97

Proposition 6.21. With the above notations, we have the formula

2 X 2 2 2 Φ = N + (|νi| − |νj| ) i6=j where ν = (ν0, . . . , νN−1) is the vector given by ν = F q. Proof. This follows by replacing in the above proof the Cauchy-Schwarz estimate by the corresponding sum of squares. More precisely, we know from the above proof that:

3 X 4 Φ = N |ξi| i P 2 On the other hand Uij = ξj−i being unitary, we have i |ξi| = 1, and so: X 4 X 2 2 1 = |ξi| + |ξi| · |ξj| i i6=j ! X 4 X 4 X 2 2 = N |ξi| − (N − 1) |ξi| − |ξi| · |ξj| i i i6=j 1 X = Φ − (|ξ |2 − |ξ |2)2 N 2 i j i6=j 2 Now by multiplying by N , this gives the formula in the statement. All this is quite interesting. As an application, in the real Hadamard matrix case, we have the following analytic reformulation of the CHC, from [18]:

N Theorem 6.22. For q ∈ T satisfying q¯i = q−i, the following quantity is real, X Φ = qiqjqkql i+j+k+l=0 and satisfies Φ ≥ N 2. The CHC states that we cannot have equality at N > 4. Proof. This follows indeed from Theorem 6.20, via the identifications from Theorem 6.15, N the parameter space in the real case being {q ∈ T |q¯i = q−i}. This is certainly quite nice, and the analytic problem might look quite elementary. However, this is not the case. In fact, we already know from section 1 that the CHC is equivalent to Ryser’s conjecture, which looks elementary as well, and is not. Following [18], let us further discuss all this. We first have:

Theorem 6.23. Write Φ = Φ0 +...+ΦN−1, with each Φi being given by the same formula as Φ, namely Φ = P qiqk , but keeping the index i ﬁxed. Then: i+k=j+l qj qk (1) The critical points of Φ are those where Φi ∈ R, for any i. (2) In the Hadamard case we have Φi = N, for any i. 98 TEO BANICA

Proof. This follows by doing some elementary computations, as follows: (1) The ﬁrst observation is that the non-constant terms in the deﬁnition of Φ involving ¯ the variable qi are the terms of the sum Ki + Ki, where: 2 X qi X X qiqk Ki = + 2 qjql qjql 2i=j+l k6=i i+k=j+l

iαi Thus if we ﬁx i and we write qi = e , we obtain: ! ∂Φ X X qiqk = 4Re i · ∂αi qjql k i+k=j+l ! X qiqk = 4Im qjql i+k=j+l

= 4Im(Φi) Now since the derivative must vanish for any i, this gives the result. (2) We ﬁrst perform the end of the Fourier computation in the proof of Theorem 6.20 above backwards, by keeping the index i ﬁxed. We obtain: X qiqk Φi = qjql i+k=j+l

1 X X qiqk = w(i−j+k−l)s N qjql s ijkl 1 X X X X = wsiq w−sjq¯ wskq w−slq¯ N i j k l s j k l 2 X si ¯ ¯ = N w qiξsξsξs s √ Here we have used the formula√ξ = F q/ N. Now by assuming that we are in the Hadamard case, we have |ξs| = 1/ N for any s, and so we obtain: √ X si ¯ ∗ Φi = N w qiξs = N Nqi(F ξ)i = Nqiq¯i = N s Thus, we have obtained the conclusion in the statement. Let us discuss now a probabilistic approach to all this. Given a compact manifold X endowed with a probability measure, and a bounded function Θ : X → [0, ∞), the maximum of this function can be recaptured via following well-known formula: Z 1/p max Θ = lim Θ(x)p dx p→∞ X HADAMARD MATRICES 99

In our case, we are rather interested in computing a minimum, and the result is: Proposition 6.24. We have the formula Z 1/p min Φ = N 3 − lim (N 3 − Φ)p dq p→∞ TN where the torus TN is endowed with its usual probability measure. Proof. This follows from the above formula, with Θ = N 3 − Φ. Observe that Θ is indeed 3 positive, because Φ is by deﬁnition a sum of N complex numbers of modulus 1. Let us restrict now the attention to the problem of computing the moments of Φ, which is more or less the same as computing those of N 3 − Φ. We have here: Proposition 6.25. The moments of Φ are given by Z p i1k1 . . . ipkp Φ dq = # is + ks = js + ls, [i1k1 . . . ipkp] = [j1l1 . . . jplp] N j1l1 . . . jplp T where the sets between brackets are by deﬁnition sets with repetition.

Proof. This is indeed clear from the formula of Φ. See [19]. Regarding now the real case, an analogue of Proposition 6.25 holds, but the combinatorics does not get any simpler. One idea in dealing with this problem is by considering the “enveloping sum”, obtained from Φ by dropping the condition i + k = j + l: X qiqk Φ˜ = qjql ijkl The point is that the moments of Φ appear as “sub-quantities” of the moments of Φ,˜ so perhaps the question to start with is to understand very well the moments of Φ.˜ ˜ P 4 And this latter problem sounds like a quite familiar one, because Φ = | i qi| . We will be back to this a bit later. For the moment, let us do some combinatorics: Proposition 6.26. We have the moment formula Z X 2p N! Φ˜ p dq = N π (N − |π|)! T π∈P (2p) 2p 2p where = , with b1, . . . , b|π| being the lengths of the blocks of π. π b1,...,b|π| Proof. Indeed, by using the same method as for Φ, we obtain: Z ˜ p i1k1 . . . ipkp Φ(q) dq = # [i1k1 . . . ipkp] = [j1l1 . . . jplp] N j1l1 . . . jplp T The sets with repetitions on the right are best counted by introducing the corresponding partitions π = ker i1k1 . . . ipkp , and this gives the formula in the statement. 100 TEO BANICA

In order to discuss now the real case, we have to slightly generalize the above result, by computing all the half-moments of Φ.e The result here is best formulated as: Proposition 6.27. We have the moment formula Z X 2p X N! | qi| dq = Cpk N (N − k)! T i k P p where Cpk = , with b1, . . . , b|π| being the lengths of the blocks of π. π∈P (p),|π|=k b1,...,b|π| Proof. This follows indeed exactly as Proposition 6.26 above, by replacing the exponent p by the exponent p/2, and by splitting the resulting sum as in the statement. Observe that the above formula basically gives the moments of Φ,˜ in the real case. Indeed, let us restrict attention to the case N = 2m. Then, for the purposes of our minimization problem we can assume that our vector is of the following form:

q = (1, q1, . . . , qm−1, 1, q¯m−1,..., q¯1) So, we are led to the following conclusion, relating the real and complex cases: m−1 Proposition 6.28. Consider the variable X = q1 + ... + qm−1 over the torus T . (1) For the complex problem at N = m − 1, we have Φe = |X|4 (2) For the real problem at N = 2m, we have Φe = |2 + X + X¯|4.

Proof. This is indeed clear from the deﬁnition of the enveloping sum Φ.e Finally, here is a random walk formulation of the problem: Proposition 6.29. The moments of Φ have the following interpretation: R p (1) First, the moments of the enveloping sum Φe count the loops of length 4p on the standard lattice ZN ⊂ RN , based at the origin. (2) R Φp counts those loops which are “piecewise balanced”, in the sense that each of the p consecutive 4-paths forming the loop satisfy i + k = j + l modulo N. Proof. The ﬁrst assertion follows from the formula in the proof of Proposition 6.27, and the second assertion follows from the formula in Proposition 6.25. This statement looks quite encouraging, but passing from (1) to (2) is quite a delicate task, because in order to interpret the condition i + k = j + l we have to label the N coordinate axes of R by elements of the cyclic group ZN , and this is a quite unfamiliar operation. In addition, in the real case the combinatorics becomes more complex due to the symmetries of the parameter space, and no further results are available so far. HADAMARD MATRICES 101

7. Bistochastic matrices In this section and the next two ones we discuss certain analytic aspects of the complex Hadamard matrices, based on the various inequalities obtained in section 1. We will extend these inequalities to the complex case, and discuss them in detail. As a ﬁrst, fundamental result, we have: Theorem 7.1. The complex Hadamard matrices, which form the manifold √ XN = MN (T) ∩ NUN can be analytically detected in two ways, as follows: √ N/2 (1) Given H ∈ √MN (T) we have | det(H)| ≤ N , with equality when H ∈ NUN . 2 (2) Given H ∈ NUN we have ||H||1 ≤ N , with equality when H ∈ MN (T). Proof. This is something that we already know in the real case, from Theorem 1.17 and Theorem 1.18 above, and the proof in the general case is similar: N (1) This follows indeed as in the real case, because if we denote by H1,...,HN ∈ T the rows of H, then we have, according to the deﬁnition of the determinant:

| det(H)| = vol < H1,...,HN > ≤ ||H || × ... × ||H || √ 1 N = ( N)N

The equality holds when H1,...,HN are pairwise orthogonal, as claimed. (2) This is something that we discussed in much detail in Proposition 6.17, Theorem 6.18 and Proposition 6.19, and which follows from Cauchy-Schwarz: !1/2 X X 2 2 ||H||1 = |Hij| ≤ N |Hij| = N ij ij

The equality case holds when |Hij| = 1 for any i, j, as claimed. We will further discuss all this in section 9 below, with a few comments on (1), and with a detailed study of the condition (2), which is something quite fruitful. Regarding now the third and last basic inequality from the real case, namely the excess estimate from Theorem 1.19, this is something of a diﬀerent nature, that we will discuss in this section, and in the next one. Let us begin with the following deﬁnition:

Deﬁnition 7.2. A complex Hadamard matrix H ∈ MN (C) is called bistochastic when the sums on all rows and all columns are equal. We denote by n o bis XN = H ∈ XN H = bistochastic the real algebraic manifold formed by such matrices. 102 TEO BANICA

The bistochastic Hadamard matrices are quite interesting objects, and include for instance all the circulant Hadamard matrices, that we discussed in section 6. Indeed, assuming that Hij = ξj−i is circulant, all rows and columns sum up to: X λ = ξi i So, let us first review the material in section 6, from this perspective. As a first and trivial remark, the Fourier matrix F2 looks better in bistochastic form: 1 1 i 1 F = ∼ = F 0 2 1 −1 1 i 2 This is something quite interesting, philosophically speaking. Indeed, we have here a new idea, namely that of studying the Hadamard matrices H ∈ MN (±1) by putting them 0 in complex bistochastic form, H ∈ MN (T), and then studying these latter matrices. We will see later on that, while certainly being viable, this idea is quite difficult to develop in practice, and is in need of some considerable preliminary work.

Let us keep now reviewing the material in section 6. According to the results there, and to the above-mentioned fact that circulant implies bistochastic, we have: Theorem 7.3. The class of bistochastic Hadamard matrices is stable under permuting rows and columns, and under taking tensor products. As examples, we have: 0 (1) The circulant and symmetric forms FN of the Fourier matrices FN . 0 (2) The bistochastic and symmetric forms FG of the Fourier matrices FG. (3) The circulant and symmetric Backelin matrices, having size MN with M|N. Proof. In this statement the claim regarding permutations of rows and columns is clear. Assuming now that H,K are bistochastic, with sums λ, µ, we have: X X X X (H ⊗ K)ia,jb = HijKab = Hij Kab = λµ ia ia i a X X X X (H ⊗ K)ia,jb = HijKab = Hij Kab = λµ jb jb j b Thus, the matrix H ⊗K is bistochastic as well. As for the assertions (1,2,3), we already know all this, from Theorem 6.6, Theorem 6.8 and Theorem 6.9 above. In the above list of examples, (2) is the key entry. Indeed, while many interesting matrices, such as the usual Fourier ones FN , can be put in circulant form, this is something quite exceptional, which does not work any longer when looking at the general Fourier matrices FG. To be more precise, when using a decomposition of type G = ZN1 ×...×ZNk , and setting F 0 = F 0 ⊗ ... ⊗ F 0 , we can only say that F 0 is bistochastic. G N1 Nk G HADAMARD MATRICES 103

As a conclusion, the bistochastic Hadamard matrices are interesting objects, deﬁnitely worth some study. So, let us develop now some general theory, for such matrices. First, we have the following elementary result:

Proposition 7.4. For an Hadamard matrix H ∈ MN (C), the following are equivalent: (1) H is bistochastic, with sums λ. (2) H is row-stochastic, with sums λ, and |λ|2 = N. Proof. Both the implications are elementary, as follows: N (1) =⇒ (2) If we denote by H1,...,HN ∈ T the rows of H, we have indeed: X N = < H1,H1 >= < H1,Hi > i X X ¯ X X ¯ = H1jHij = H1j Hij i j j i X ¯ ¯ 2 = H1j · λ = λ · λ = |λ| j

N (2) =⇒ (1) Consider the all-one vector ξ = (1)i ∈ C . The fact that H is row- stochastic with sums λ, respectively column-stochastic with sums λ, reads: X X Hij = λ, ∀i ⇐⇒ Hijξj = λξi, ∀i ⇐⇒ Hξ = λξ j j X X t Hij = λ, ∀j ⇐⇒ Hijξi = λξj, ∀j ⇐⇒ H ξ = λξ i j We must prove that the ﬁrst condition implies the second one, provided that the row sum λ satisﬁes |λ|2 = N. But this follows from the following computation: Hξ = λξ =⇒ H∗Hξ = λH∗ξ =⇒ N 2ξ = λH∗ξ =⇒ N 2ξ = λH¯ tξ =⇒ Htξ = λξ

Thus, we have proved both the implications, and we are done. Here is another basic result, that we will need as well in what follows:

Proposition 7.5. For a complex Hadamard matrix H ∈ MN (C), and a number λ ∈ C satisfying |λ|2 = N, the following are equivalent: (1) We have H ∼ H0, with H0 being bistochastic, with sums λ. N (2) Kij = aibjHij is bistochastic with sums λ, for some a, b ∈ T . (3) The equation Hb = λa¯ has solutions a, b ∈ TN . 104 TEO BANICA

Proof. Once again, this is an elementary result, the proof being as follows: (1) ⇐⇒ (2) Since the permutations of the rows and columns preserve the bistochas- ticity condition, the equivalence H ∼ H0 that we are looking for can be assumed to come only from multiplying the rows and columns by numbers in T. Thus, we are looking for scalars ai, bj ∈ T such that Kij = aibjHij is bistochastic with sums λ, as claimed. (2) ⇐⇒ (3) The row sums of the matrix Kij = aibjHij are given by: X X Kij = aibjHij = ai(Hb)i j j Thus K is row-stochastic with sums λ precisely when Hb = λa¯, and by using the equivalence in Proposition 7.4, we obtain the result. Finally, here is an extension of the excess inequality from Theorem 1.19 above:

Theorem 7.6. For a complex Hadamard matrix H ∈ MN (C), the excess, X E(H) = Hij ij √ satisﬁes |E(H)| ≤ N N, with equality if and only if H is bistochastic.

N Proof. In terms of the all-one vector ξ = (1)i ∈ C , we have: X X ¯ X ¯ E(H) = Hij = Hijξjξi = (Hξ)iξi =< Hξ, ξ > ij ij i √ Now by using the Cauchy-Schwarz inequality, along with the fact that U = H/ N is unitary, and hence of norm 1, we obtain, as claimed: √ |E(H)| ≤ ||Hξ|| · ||ξ|| ≤ ||H|| · ||ξ||2 = N N Regarding now the equality case, this requires the vectors Hξ, ξ to be proportional, and so our matrix H to be row-stochastic. Moreover, if we assume Hξ = λξ, the above 2 computation gives |λ| = N, and by Proposition 7.4, we obtain the result. The above estimate is potentially quite useful, because it allows us to analytically locate bis the bistochastic Hadamard manifold XN inside the whole Hadamard manifold XN , a bit in the spirit of the two analytic methods in Theorem 7.1. We will be back to this later, with a number of probabilistic results on the subject.

Let us go back now to the fundamental question, which already appeared several times in the above, of putting an arbitrary Hadamard matrix in bistochastic form. As already explained, we are interested in solving this question in general, and in particular in the real case, with potential complex reformulations of the HC and CHC at stake. What we know so far on this subject can be summarized as follows: HADAMARD MATRICES 105

Proposition 7.7. An Hadamard matrix H ∈ MN (C) can be put in bistochastic form when one of the following conditions is satisfied: √ N (1) The equations |Ha|i = N, with i = 1,...,N√ , have solutions a ∈ T . (2) The quantity |E| attains its maximum N N over the equivalence class of H. Proof. This follows indeed from Proposition 7.4 and Proposition 7.5 above. Thus, we have two approaches to the problem, one algebraic, and one analytic. Let us first discuss the algebraic approach, coming from (1) above. What we have there is a certain system of N equations, having as unknowns N real variables, namely the phases of a1, . . . , aN . This system is highly non-linear, but can be solved, however, via a certain non-explicit method, as explained by Idel and Wolf in [57]. In order to discuss this material, which is quite advanced, let us begin with some preliminaries. The complex projective space appears by definition as follows: P N−1 = ( N − {0}) < x = λy > C C Inside this projective space, we have the Clifford torus, constructed as follows: n o N−1 = (z , . . . , z ) ∈ P N−1 |z | = ... = |z | T 1 N C 1 N With these conventions, we have the following result, from [57]:

Proposition 7.8. For a unitary matrix U ∈ UN , the following are equivalent: 0 (1) There exist L, R ∈ UN diagonal such that U = LUR is bistochastic. (2) The standard torus TN ⊂ CN satisfies TN ∩ UTN 6= ∅. (3) The Clifford torus N−1 ⊂ P N−1 satisfies N−1 ∩ U N−1 6= ∅. T C T T Proof. These equivalences are all elementary, as follows: (1) =⇒ (2) Assuming that U 0 = LUR is bistochastic, which in terms of the all-1 vector ξ means U 0ξ = ξ, if we set f = Rξ ∈ TN we have: 0 0 N Uf = LU¯ Rf¯ = LU¯ ξ = Lξ¯ ∈ T Thus we have Uf ∈ TN ∩ UTN , which gives the conclusion. (2) =⇒ (1) Given g ∈ TN ∩ UTN , we can define R,L as follows: ¯ R = diag(g1, . . . , gN ) , L = diag((Ug)1,..., (Ug)N ) We have then Rξ = g and Lξ¯ = Ug, and so U 0 = LUR is bistochastic, because: U 0ξ = LURξ = LUg = ξ (2) =⇒ (3) This is clear, because N−1 ⊂ P N−1 appears as the projective image of T C TN ⊂ CN , and so TN−1 ∩ UTN−1 appears as the projective image of TN ∩ UTN . (3) =⇒ (2) We have indeed the following equivalence: N−1 N−1 N N T ∩ UT 6= ∅ ⇐⇒ ∃λ 6= 0, λT ∩ UT 6= ∅ But U ∈ UN implies |λ| = 1, and this gives the result. 106 TEO BANICA

The point now is that the condition (3) above is something familiar in symplectic geometry, and known to hold for any U ∈ UN . Thus, following [57], we have:

Theorem 7.9. Any unitary matrix U ∈ UN can be put in bistochastic form, U 0 = LUR with L, R ∈ UN being both diagonal, via a certain non-explicit method.

Proof. As already mentioned, the condition TN−1 ∩ UTN−1 6= ∅ in Proposition 7.8 (3) is something quite natural in symplectic geometry. To be more precise, N−1 ⊂ P N−1 is T C a Lagrangian submanifold, TN−1 → UTN−1 is a Hamiltonian isotopy, and a result from [29], [35] states that TN−1 cannot be displaced from itself via a Hamiltonian isotopy. Thus, the results in [29], [35] tells us that TN−1 ∩ UTN−1 6= ∅ holds indeed, for any U ∈ UN . We therefore obtain the result, via Proposition 7.8. See [57]. In relation now with our Hadamard matrix questions, we have: Theorem 7.10. Any complex Hadamard matrix can be put in bistochastic form, up to the standard equivalence relations for such matrices. √ Proof. This√ follows indeed from Theorem 7.9, because if H = NU is Hadamard then so is H0 = NU 0, and with the remark that, in what regards the equivalence relation, we just need the multiplication of the rows and columns by scalars in T. All this is extremely interesting, but unfortunately, not explicit. As explained in [57], the various technical results from [29], [35] show that in the generic, “transverse” situation, N−1 there are at least 2 ways of putting a unitary matrix U ∈ UN in bistochastic form, and this modulo the obvious transformation U → zU, with |z| = 1. Summarizing, the question of explicitely putting the Hadamard matrices H ∈ MN (C) in bistochastic form remains open, and open as well is the question of ﬁnding a simpler proof for the fact that this can be done indeed, without using [29], [35].

Regarding this latter question, a possible approach comes from the excess result from Theorem 7.6 above. Indeed, in view of the remark√ there, it is enough to show that the law of |E| over the equivalence class of H has N N as upper support bound. In order to comment on this, let us ﬁrst formulate:

Deﬁnition 7.11. The glow of H ∈ MN (C) is the measure µ ∈ P(C) given by: ! Z Z X ϕ(x)dµ(x) = ϕ aibjHij d(a, b) N N C T ×T ij P That is, the glow is the law of E = ij Hij, over the equivalence class of H. HADAMARD MATRICES 107

In this deﬁnition H can be any complex matrix, but the equivalence relation is the one for the complex Hadamard matrices. To be more precise, let us call two matrices H,K ∈ MN (C) equivalent if one can pass from one to the other by permuting rows and columns, or by multiplying the rows and columns by numbers in T. Now since P permuting rows and columns does not change the quantity E = ij Hij, we can restrict N N attention from the full equivalence group G = (SN o T ) × (SN o T ) to the smaller group G0 = TN × TN , and we obtain the measure µ in Deﬁnition 7.11.

As in the real case, the terminology comes from a picture of the following type, with the stars ∗ representing the entries of our matrix, and with the switches being supposed now to be continuous, randomly changing the phases of the concerned entries:

→ ∗ ∗ ∗ ∗ → ∗ ∗ ∗ ∗ → ∗ ∗ ∗ ∗ → ∗ ∗ ∗ ∗

↑ ↑ ↑ ↑

In short, what we have here is a complex generalization of the Gale-Berlekamp game [50], [82], and this is where the main motivation for studying the glow comes from.

We are in fact interested in computing a real measure, because we have:

Proposition 7.12. The laws µ, µ+ of E, |E| over the torus TN × TN are related by µ = ε × µ+ where × is the multiplicative convolution, and ε is the uniform measure on T. Proof. We have E(λH) = λE(H) for any λ ∈ T, and so µ = law(E) is invariant under the action of T. Thus µ must decompose as µ = ε × µ+, where µ+ is a certain probability + measure on [0, ∞), and this measure µ is the measure in the statement. In particular, we can see from the above result that the glow is invariant under rotations. With this observation made, we can formulate the following result:

Theorem√ 7.13. The glow of any Hadamard matrix H ∈ MN (C), or more generally of any H ∈ NUN , satisﬁes the following conditions, where D is the unit disk, √ √ N N T ⊂ supp(µ) ⊂ N N D with the inclusion on the right coming from Cauchy-Schwarz, and with the inclusion on the left corresponding to the fact that H can be put in bistochastic form. 108 TEO BANICA

Proof. In this statement the inclusion on the right comes indeed from Cauchy-Schwarz, as explained in the proof of Theorem 7.6 above, with the√ remark that the computation there only uses the fact that the rescaled matrix U = H/ N is unitary. Regarding now the inclusion on the left, we know from Theorem 7.9 that H can be put in bistochastic form. According to Proposition 7.7, this tells us that we have: √ N N T ∩ supp(µ) 6= ∅ Now by using the rotational invariance of the√ glow, and hence of its support, coming from Proposition 7.12, we obtain from this N N T ⊂ supp(µ), as claimed. The challenging question is that of obtaining a proof of the above result by using probabilistic techniques. Indeed, as explained at the end of section 6 above, the support of a measure can be recaptured from the moments, simply by computing a limit. Thus, knowing the moments of the glow well enough would solve the problem. Regarding the moments of the glow, the formula is as follows:

Proposition 7.14. For H ∈ MN (T) the even moments of |E| are given by Z X Hi1j1 ...Hipjp |E|2p = N N Hk l ...Hk l T ×T [i]=[k],[j]=[l] 1 1 p p where the sets between brackets are by deﬁnition sets with repetition. Proof. We have indeed the following computation: Z Z 2p 2p X |E| = Hijaibj N N N N T ×T T ×T ij Z !p X Hij aibj = · N N Hkl akbl T ×T ijkl Z Z X Hi1j1 ...Hipjp ai1 . . . aip bj1 . . . bjp = Hk l ...Hk l N ak . . . ak N bl . . . bl ijkl 1 1 p p T 1 p T 1 p

Now since the integrals at right equal respectively the Kronecker symbols δ[i],[k] and δ[j],[l], we are led to the formula in the statement. With this formula in hand, the main result, regarding the fact that the complex Hada- mard matrices can be put in bistochastic form, reformulates as follows:

Theorem 7.15. For a complex Hadamard matrix H ∈ MN (T) we have  1/p X Hi1j1 ...Hipjp 3 lim   = N p→∞ Hk l ...Hk l [i]=[k],[j]=[l] 1 1 p p coming from the fact that H can be put in bistochastic form. HADAMARD MATRICES 109

Proof. This follows from the well-known fact that the maximum of a bounded function Θ: X → [0, ∞) can be recaptured via following formula: Z 1/p max(Θ) = lim Θ(x)p dx p→∞ X With X = TN × TN and with Θ = |E|2, we conclude that the limit in the statement is the square of the upper bound of the glow. But, according to Theorem 7.13, this upper 3 bound is known to be ≤ N by Cauchy-Schwarz, and the equality holds by [57]. To conclude now, the challenging question is that of finding a direct proof for Theorem 7.15. All this would provide an alternative aproach to the results in [57], which would be of course still not explicit, but which would use at least some more familiar tools. We will discuss such questions in section 9 below, with the remark however that the problems at N ∈ N fixed being quite difficult, we will do a N → ∞ study only.

Getting away now from these difficult questions, we have nothing concrete so far, besides the list of examples from Theorem 7.3, coming from the circulant matrix considerations in section 6. So, our purpose will be that of extending that list. A first natural question is that of looking at the Butson matrix case. We have here the following result, extending the finding from Proposition 7.4 above:

Proposition 7.16. Assuming that HN (l) contains a bistochastic matrix, the equations

a0 + a1 + ... + al−1 = N l−1 2 |a0 + a1w + ... + al−1w | = N must have solutions, over the positive integers. Proof. This is a reformulation of the equality |λ|2 = N, from Proposition 7.4 above. 2πi/l i Indeed, if we set w = e , and we denote by ai ∈ N the number of w entries appearing in the ﬁrst row of our matrix, then the row sum of the matrix is given by: l−1 λ = a0 + a1w + ... + al−1w Thus, we obtain the system of equations in the statement. The point now is that, in practice, we are led precisely to the Turyn obstructions from section 6 above. At very small values of l, the obstructions are as follows:

Theorem 7.17. Assuming that HN (l) contains a bistochastic matrix, the following equations must have solutions, over the integers: (1) l = 2: 4n2 = N. (2) l = 3: x2 + y2 + z2 = 2N, with x + y + z = 0. (3) l = 4: a2 + b2 = N. 110 TEO BANICA

Proof. This follows indeed from the results that we have: (1) This is something well-known, which follows from Proposition 7.16. (2) This is best viewed by using Proposition 7.16, and the following formula, that we already know, from section 3 above: 1 |a + bw + cw2|2 = [(a − b)2 + (b − c)2 + (c − a)2] 2 At the level of the concrete obstructions, we must have for instance 56 |N. Indeed, this follows as in the proof of the de Launey obstruction for HN (3) with 5|N. 2 2 2 (3) This follows again from Proposition 7.16, and from |a + ib| = a + b . As a conclusion, nothing much interesting is going on in the Butson matrix case, with various arithmetic obstructions, that we partly already met, appearing here. See [63].

In order to reach, however, to a number of positive results, beyond those in Theorem 7.3, we can investigate various special classes of matrices, such as the Dit¸˘aproducts. In order to formulate our results, we will use the following notion:

Deﬁnition 7.18. We say that a complex Hadamard√ matrix H ∈ MN (C) is in “almost bistochastic form” when all the row sums belong to N · T. Observe that, assuming that this condition holds, the matrix H can be put in bistochastic form, just by multiplying its rows by suitable numbers from T. We will be particularly interested here in the special situation where the aﬃne defor- q mations H ∈ MN (C) of a given complex Hadamard matrix H ∈ MN (C) can be put in almost bistochastic form, independently of the value of the parameter q. For the simplest deformations, namely those of F2 ⊗ F2, this is indeed the case: p q Proposition 7.19. The deformations of F2 ⊗ F2, with parameter matrix Q = (r s), p q p q  p −q p −q F2 ⊗Q F2 =   r s −r −s r −s −r s can be put in almost bistochastic form, independently of the value of Q. Proof. By multiplying the columns of the matrix in the statement with 1, 1, −1, 1 respectively, we obtain the following matrix: p q −p q  00 p −q −p −q F2 ⊗ F2 =   Q r s r −s r −s r s The row sums of this matrix being 2q, −2q, 2r, 2r ∈ 2T, we are done. HADAMARD MATRICES 111

00 We will see later on that the above matrix F2 ⊗Q F2 is equivalent to a certain matrix 0 0 F2 ⊗ F2, which looks a bit more complicated, but is part of a series FN ⊗ FN . Now back to the general case, we have the following result:

Theorem 7.20. A deformed tensor product H ⊗Q K can be put in bistochastic form when i there exist numbers xa ∈ T such that with ∗ i (K x )b Gib = Qib √ ∗ we have |(H G)ib| = MN, for any i, b.

Proof. The deformed tensor product L = H ⊗Q K is given by Lia,jb = QibHijKab. By multiplying the columns by scalars Rjb ∈ T, this matrix becomes: 0 Lia,jb = RjbQibHijKab The row sums of this matrix are given by:

0 X Sia = RjbQibHijKab jb X X = KabQib HijRjb b j X = KabQib(HR)ib b i In terms of the variables Cb = Qib(HR)ib, these rows sums are given by: 0 X i i Sia = KabCb = (KC )a b

Thus H ⊗Q K can be put in bistochastic form when we can ﬁnd scalars Rjb ∈ T and i i xa ∈ T such that, with Cb = Qib(HR)ib, the following condition is satisﬁed: √ i i (KC )a = MNxa , ∀i, a √ But this condition is equivalent to KCi = MNxi for any i, and by multiplying to the left by the adjoint matrix K∗, we are led to the following condition: √ √ NCi = MK∗xi , ∀i i Now by recalling that Cb = Qib(HR)ib, this condition is equivalent to: √ √ ∗ i NQib(HR)ib = M(K x )b , ∀i, b ∗ i With Gib = (K x )b/Qib as in the statement, this condition reads: √ √ N(HR)ib = MGib , ∀i, b 112 TEO BANICA √ √ But this condition is equivalent to NHR = MG, and by multiplying to the left by the adjoint matrix H∗, we are led to the following condition: √ MNR = H∗G

Thus, we have obtained the condition in the statement.

As an illustration for this result, assume that H,K can be put in bistochastic form, by M N i using vectors y ∈ T , z ∈ T . If we set xa = yiza, with Q = 1 we have:

∗ i ∗ ∗ Gib = (K x )b = [K (yiz)]b = yi(K z)b

We therefore obtain the following formula:

∗ X ∗ (H G)ib = (H )ijGjb j X ∗ ∗ = (H )ijyj(K z)b j ∗ ∗ = (H y)i(K z)b

Thus the usual tensor product H ⊗ K can be put in bistochastic form as well. In the case H = FM the equations simplify, and we have:

Proposition 7.21. A deformed tensor product FM ⊗Q K can be put in bistochastic form i when there exist numbers xa ∈ T such that with

∗ i (K x )b Gib = Qib we have the following formulae, with l being taken modulo M:

X ¯ GjbGj+l,b = MNδl,0 , ∀l, b j

2 2 Moreover, the M×N matrix |Gjb| is row-stochastic with sums N , and the l = 0 equations state that this matrix must be column-stochastic, with sums MN.

Proof. With notations from Theorem 7.20, and with w = e2πi/M , we have:

∗ X −ij (H G)ib = w Gjb j HADAMARD MATRICES 113

The absolute value of this number can be computed as follows:

∗ 2 X i(k−j) ¯ |(H G)ib| = w GjbGkb jk X il ¯ = w GjbGj+l,b jl X il X ¯ = w GjbGj+l,b l j

b If we denote by vl the sum on the right, we obtain:

∗ 2 X il b b |(H G)ib| = w vl = (FM v )i l √ M ∗ Now if we denote by ξ the all-one vector in C , the condition |(H G)ib| = MN for any i, b found in Theorem 7.20 above reformulates as follows: F M vb = MNξ , ∀b

∗ By multiplying to the left by FM /M, this condition is equivalent to: MN 0 b ∗   v = NF ξ =  .  , ∀b M  .  0

b b Let us examine the ﬁrst equation, v0 = MN. By deﬁnition of vl , we have: b X ¯ X 2 v0 = GjbGjb = |Gjb| j j

∗ j Now recall from Theorem 7.20 that we√ have Gjb = (K x )b/Qjb, for certain numbers j ∗ xb ∈ T. Since we have Qjb ∈ T and K / N ∈ UN , we obtain:

X 2 X ∗ j 2 |Gjb| = |(K x )b| b b ∗ j 2 = ||K x ||2 j 2 = N||x ||2 = N 2

2 2 Thus the M × N matrix |Gjb| is row-stochastic, with sums N , and our equations b v0 = MN for any b state that this matrix must be column-stochastic, with sums MN. 114 TEO BANICA

b Regarding now the other equations that we found, namely vl = 0 for l 6= 0, by deﬁnition b of vl and of the variables Gjb, these state that we must have: X ¯ GjbGj+l,b = 0 , ∀l 6= 0, ∀b j

Thus, we are led to the conditions in the statement. As an illustration for this result, let us go back to the Q = 1 situation, explained after ∗ Theorem 7.20. By using the formula Gib = yi(K z)b there, we have:

X ¯ X ∗ ∗ GjbGj+l,b = yj(K z)b yj+l(K z)b j j X yj = |(K∗z) |2 b y j j+l

= M · Nδl,0

Thus, if K can be put in bistochastic form, then so can be put FM ⊗ K. 0 As a second illustration, let us go back to the matrices F2 ⊗Q F2 from the proof of Proposition 7.19 above. The vector of the row sums being S = (2q, −2q, 2r, 2r), we have x = (q, −q, r, r), and so we obtain the following formulae for the entries of G: 1 1 q 0 1 −1 −q 2q b b G0b = = Q0b Q0b 1 1 r 2r 1 −1 r 0 b b G1b = = Q1b Q1b Thus, in this case the matrix G is as follows, independently of Q: 0 2 G = 2 0 In particular, we see that the conditions in Proposition 7.21 are satisﬁed. As a main application now, we have the following result:

Theorem 7.22. The Dit¸˘adeformations FN ⊗Q FN can be put in almost bistochastic form, independently of the value of the parameter matrix Q ∈ MN (T). 2πi/N Proof. We use Proposition 7.21 above, with M = N, and with K = FN . Let w = e , and consider the vectors xi ∈ TN given by: i (i−1)a x = (w )a HADAMARD MATRICES 115

∗ i Since K K = N1N , and x are the column vectors of K, shifted by 1, we have:  0  N  0   0   0   0  ∗ 0  .  ∗ 1  .  ∗ N−1  .  K x =  .  ,K x =  .  ,...,K x =  .         0   0  N N 0 0 ∗ i We conclude that we have (K x )b = Nδi−1,b, and so the matrix G is given by:

Nδi−1,b Gib = Qib With this formula in hand, the sums in Proposition 7.21 are given by:

X X Nδj−1,b Nδj+l−1,b G G¯ = · jb j+l,b Q Q j j jb j+l,b In the case l 6= 0 we clearly get 0, because the products of Kronecker symbols are 0. In 2 the case l = 0 the denominators are |Qjb| = 1, and we obtain: X ¯ 2 X 2 GjbGjb = N δj−1,b = N j j Thus, the conditions in Proposition 7.21 are satisﬁed, and we obtain the result. Here is an equivalent formulation of the above result:

0 Theorem 7.23. The matrix FN ⊗Q FN , with Q ∈ MN (T), defined by ij+ab 0 w Qib (FN ⊗Q FN )ia,jb = bj+j · w Qb+1,b 2πi/N where w = e is almost bistochastic, and equivalent to FN ⊗Q FN . Proof. Our claim is that this is the matrix constructed in the proof of Theorem 7.22. Indeed, let us first go back to the proof of Theorem 7.20. In the case M = N and H = K = FN , the Dit¸˘adeformation L = H ⊗Q K studied there is given by: ij+ab Lia,jb = QibHijKab = w Qib As explained in the proof of Theorem 7.22, if the conditions in the statement there are 0 satisfied, then the matrix Lia,jb = RjbLia,jb is almost bistochastic, where: √ MN · R = H∗G

In our case now, M = N and H = K = FN , we know from the proof of Proposition 7.21 that the choice of G which makes work Theorem 7.22 is as follows: Nδi−1,b Gib = Qib 116 TEO BANICA

With this formula in hand, we can compute the matrix R, as follows: 1 R = (H∗G) jb N jb 1 X = w−ijG N ib i X δi−1,b = wij · Q i ib w−(b+1)j = Qb+1,b

Thus, the modiﬁed version of FN ⊗Q FN which is almost bistochastic is given by: 0 Lia,jb = RjbLia,jb −(b+1)j w ij+ab = · w Qib Qb+1,b ij+ab w Qib = bj+j · w Qb+1,b Thus we have obtained the formula in the statement, and we are done. As an illustration, let us work out the case N = 2. Here we have w = −1, and with p q p s Q = (r s), and then with u = r , v = q , we obtain the following matrix:  p q p q    r q − r q u 1 −u 1 p q p q  r − q − r − q  u −1 −u −1 F2 ⊗Q F2 =  r s r s  =    r q r − q  1 v 1 −v r s r s 1 −v 1 v r − q r q Observe that this matrix is indeed almost bistochastic, with row sums 2, −2, 2, 2. It is quite unclear on how to get beyond these results. An interesting question here would be probably that of focusing on the real case, and see if the Hadamard matrices there, H ∈ MN (±1), can be put or not in bistochastic form, in an explicit way. This is certainly true for the Walsh matrices, but for the other basic examples, such as the Paley or the Williamson matrices, no results seem to be known so far. Having such a theory would be potentially very interesting, with a complex reformulation of the HC and of the other real Hadamard questions at stake. HADAMARD MATRICES 117

8. Glow computations We discuss here the computation of the glow, in the N → ∞ limit. We have seen in section 1 that, in what concerns the real glow of the real Hadamard matrices, with N → ∞ we obtain a real Gaussian measure. Our main purpose here will be that of establishing a similar result in the complex case, stating that for the complex glow of the complex Hadamard matrices, with N → ∞ we obtain a complex Gaussian measure. The computations in the complex case are considerably more involved than those in the real case, and will require a lot of combinatorics. Also, we will investigate the problem of getting beyond the N → ∞ limiting result, with formulae at order 1, 2, 3, 4. As a ﬁrst motivation for all this, we have the Gale-Berlekamp game [50], [82]. Another motivation comes from the questions regarding the bistochastic matrices, in relation with [57], explained in section 7. Finally, we have the question of connecting and computing the defect, and other invariants of the Hadamard matrices, in terms of the glow.

Let us begin by reviewing the few theoretical things that we know about the glow, from section 7 above. The main results there can be summarized as follows:

Theorem 8.1. The glow of H ∈ MN (C), which is the law µ ∈ P(C) of the excess X E = Hij ij over the Hadamard equivalence class of H, has the following properties: (1) µ = ε × µ+, where µ+ = law(|E|). (2) µ is invariant√ under rotations. √ (3) H ∈ √NUN implies supp(µ) ⊂√N N D. (4) H ∈ NUN implies as well N N T ⊂ supp(µ). Proof. We already know all this from section 7, the idea being as follows: (1) This follows by using H → zH with |z| = 1, as explained in Proposition 7.12. (2) This follows from (1), the convolution with ε bringing the invariance. (3) This folllows from Cauchy-Schwarz, as explained in Theorem 7.13. (4) This is something highly non-trivial, coming from [57]. In what follows we will be mainly interested in the Hadamard matrix case, but since the computations here are quite diﬃcult, let us begin our study with other matrices. It is convenient to normalize our matrices, by assuming that the corresponding 2-norm qP 2 ||H||2 = |Hij| takes the value ||H||2 = N. Note that this is always the case with ij √ the Hadamard matrices, and more generally with the matrices H ∈ NUN . We recall that the complex Gaussian distribution C is the law of z = √1 (x + iy), where 2 x, y are independent standard Gaussian variables. In order to detect this distribution, we can use the moment method, and the well-known formula E(|z|2p) = p!. 118 TEO BANICA

Finally, we use the symbol ∼ to denote an equality of distributions. We have: Proposition 8.2. We have the following computations: √ √ N (1) For the rescaled identity IeN = NIN we have E ∼ N(q1 +...+qN ), with q ∈ T random. With N → ∞ we have E/N ∼ C. (2) For the ﬂat matrix JN = (1)ij we have E ∼ (a1 + ... + aN )(b1 + ... + bN ), with (a, b) ∈ TN × TN random. With N → ∞ we have E/N ∼ C × C.

Proof. We use Theorem√ 8.1, and the moment method: P N (1) Here we have E = N i aibi, with a, b ∈ T random, and with qi = aibi this gives the first assertion. Let us estimate now the moments of |E|2. We have: Z Z 2p p 2p |E| = N |q1 + ... + qN | dq TN ×TN TN Z X qi1 . . . qip = N p dq N q . . . q T ij j1 jp n o p p p = N # (i, j) ∈ {1,...,N} × {1,...,N} [i1, . . . , ip] = [j1, . . . , jp] ' N p · p!N(N − 1) ... (N − p + 1) ' N p · p!N p = p!N 2p Here, and in what follows, the sets between brackets are by defintion sets with repetition, and the middle estimate comes from the fact that, with N → ∞, only the multi-indices i = (i1, . . . , ip) having distinct entries contribute. But this gives the result. P P P (2) Here we have E = ij aibj = i ai j bj, and this gives the first assertion. Now N P P since a, b ∈ T are independent, so are the quantities i ai, j bj, so we have: Z Z 2 2p 2p p 2 |E| = |q1 + ... + qN | dq ' (p!N ) TN ×TN TN Here we have used the estimate in the proof of (1), and this gives the result. As a first conclusion, the glow is intimately related to the basic hypertoral law, namely N that of q1 + ... + qN , with q ∈ T random. Observe that at N = 1 this hypertoral law is simply δ1, and that at N = 2 we obtain the following law: law|1 + q| = lawp(1 + eit)(1 + e−it) √ = law 2 + 2 cos t t = law 2 cos 2 HADAMARD MATRICES 119 P In general, the law of qi is known to be related to the Pólya random walk [78]. Also, as explained for instance in section 6, the moments of this law are: Z 2p X p N! |q1 + ... + qN | dq = N π (N − |π|)! T π∈P (p)

As a second conclusion, even under the normalization ||H||2 = N, the glow can behave quite diﬀerently in the N → ∞ limit. So, let us restrict now the attention to the Hadamard matrices. At N = 2 we only have F2 to be invesigated, the result being:

Proposition 8.3. For the Fourier matrix F2 we have |E|2 = 4 + 2Re(α − β) for certain variables α, β ∈ T which are uniform, and independent.

Proof. The matrix that we interested in, namely the Fourier matrix F2 altered by a vertical switching vector (a, b) and an horizontal switching vector (c, d), is:

ac ad F = e2 bc −bd

With this notation, we have the following formula: ad bc bd ac |E|2 = |ac + ad + bc − bd|2 = 4 + + − − bc ad ac bd ad bd For proving that α = bc and β = ac are independent, we use the moment method: Z p q Z Z Z Z ad bd p−q q−p −p−q p+q = a b c d = δp,q,0 T4 bc ac T T T T Thus α, β are indeed independent, and we are done. Observe that law(|E|2), and hence law(E), is uniquely determined by the above result. It is possible of course to derive from this some more concrete formulae, but let us look instead at the case N = 3. Here the matrix that we are interested in is: ad ae af  2 Fe3 = bd wbe w bf cd w2ce wcf

Thus, we would like to compute the law of the following quantity:

|E| = |ad + ae + af + bd + wbe + w2bf + cd + w2ce + wcf| 120 TEO BANICA

The problem is that when trying to compute |E|2, the terms won’t cancel much. More 2 2 precisely, we have |E| = 9 + C0 + C1w + C2w , where C0,C1,C2 are as follows: ae ae af af bd bd be bf cd cd ce cf C = + + + + + + + + + + + 0 bd cd bd cd ae af cf ce ae af bf be ad ad ae af bd be be be cd cf cf cf C = + + + + + + + + + + + 1 bf ce bf ce ce ad af cd bf ad ae bd ad ad ae af bd bf bf bf cd ce ce ce C = + + + + + + + + + + + 2 be cf cf be cf ad ae cd be ad af bd

In short, all this leads nowhere, and the exact study stops at F2.

In general now, one idea is that of using Bernoulli-type variables coming from the row sums, as in the real case. We have here the following result:

Theorem 8.4. The glow of H ∈ MN (C) is given by the formula Z law(E) = B((Ha)1,..., (Ha)N ) a∈TN P N where B(c1, . . . , cN ) = law( i λici), with λ ∈ T random. Proof. This is indeed clear from the formula E =< a, Hb >, because when a ∈ TN is ﬁxed, E follows the law B((Ha)1,..., (Ha)N ) in the statement.

Observe that we can write B(c1, . . . , cN ) = ε × β(|c1|,..., |cN |), where the measure P β(r1, . . . , rN ) ∈ P(R+) with r1, . . . , rN ≥ 0 is given by β(r1, . . . , rN ) = law| i λiri|. Regarding now the computation of β, we have: s X λi β(r , . . . , r ) = law · r r 1 N λ i j ij j Consider now the following variable, which is easily seen, for instance by using the moment method, to be uniform over the projective torus TN−1 = TN /T: λ1 λ2 λN (µ1, µ2, . . . , µN ) = , ,..., λ2 λ3 λ1

Now since we have λi/λj = µiµi+1 . . . µj, with the convention µi . . . µj = µj . . . µi for i > j, this gives the following formula, with µ ∈ TN−1 random: sX β(r1, . . . , rN ) = law µiµi+1 . . . µj · rirj ij It is possible to further study the laws β by using this formula. However, in practice, it is more convenient to use the complex measures B from Theorem 8.4. HADAMARD MATRICES 121

Let us end these preliminaries with a discussion of the “arithmetic” version of the problem, which makes the link with the Gale-Berlekamp switching game [50], [82] and with the work in section 1. We have the following unifying formalism:

Deﬁnition 8.5. Given H ∈ MN (C) and s ∈ N ∪ {∞}, we deﬁne µs ∈ P(C) by ! Z Z X ϕ(x)dµs(x) = ϕ aibjHij d(a, b) N N C Zs ×Zs ij where Zs ⊂ T is the group of the s-roots of unity, with the convention Z∞ = T. Observe that at s = ∞ we obtain the measure in Theorem 8.1. Also, at s = 2 and for a usual Hadamard matrix, H ∈ MN (±1), we obtain the measure from section 1. Observe that for H ∈ MN (±1), knowing µ2 is the same as knowing the statistics of the number of one entries, |1 ∈ H|. This follows indeed from the following formula:

X 2 Hij = |1 ∈ H| − | − 1 ∈ H| = 2|1 ∈ H| − N ij More generally, at s = p prime, we have the following result:

Theorem 8.6. When s is prime and H ∈ MN (Zs), the statistics of the number of one P entries, |1 ∈ H|, can be recovered from that of the total sum, E = ij Hij. n Proof. The problem here is of vectorial nature, so given V ∈ Zs , we would like to compare P the quantities |1 ∈ V | and Vi. Let us write, up to permutations: V = (1 ... 1 w . . . w ...... ws−1 . . . ws−1) | {z } | {z } | {z } a0 a1 as−1 P s−1 We have then |1 ∈ V | = a0 and Vi = a0 + a1w + ... + as−1w , and we also know that a0 + a1 + ... + as−1 = n. Now when s is prime, the only ambiguity in recovering a0 s−1 s−1 from a0 + a1w + ... + as−1w can come from 1 + w + ... + w = 0. But since the sum of the numbers ai is ﬁxed, a0 + a1 + ... + as−1 = n, this ambiguity dissapears. As an illustration, at s = 3 we can write E = a + bw + cw2, and we have: b + c n − a 3a − n Re(E) = a − = a − = 2 2 2 At s = 5, however, is it now clear how to explicitely solve the problem. In fact, the problem of ﬁnding an explicit formula is related to the question of relating the complex probability measures µs = law(E) for a given matrix H ∈ MN (T), at various values of s ∈ N ∪ {∞}. Note that Theorem 8.6 tells us that for a matrix H ∈ MN (Zs) with s prime, the measure µ2 can be recaptured from the knowledge of µs.

Let us investigate now the glow of the complex Hadamard matrices, by using the moment method. We use the moment formula from section 7, namely: 122 TEO BANICA

Proposition 8.7. For H ∈ MN (T) the even moments of |E| are given by Z X Hi1j1 ...Hipjp |E|2p = N N Hk l ...Hk l T ×T [i]=[k],[j]=[l] 1 1 p p where the sets between brackets are by deﬁnition sets with repetition. P Proof. As explained in section 7, with E = ij Hijaibj we obtain: Z Z !p X Hij aibj |E|2p = · N N N N Hkl akbl T ×T T ×T ijkl Z Z X Hi1j1 ...Hipjp ai1 . . . aip bj1 . . . bjp = Hk l ...Hk l N ak . . . ak N bl . . . bl ijkl 1 1 p p T 1 p T 1 p

The integrals on the right being δ[i],[k] and δ[j],[l], we obtain the result. As a ﬁrst application, let us investigate the tensor products. We have: Proposition 8.8. The even moments of |E| for L = H ⊗ K are given by Z X Hi1j1 ...Hipjp Ka1b1 ...Kapbp |E|2p = · NM NM Hk l ...Hk l Kc d ...Kc d T ×T [ia]=[kc],[jb]=[ld] 1 1 p p 1 1 p p where the sets between brackets are as usual sets with repetition. Proof. With L = H ⊗ K, the formula in Proposition 8.7 reads: Z X Li1a1,j1b1 ...Lipap,jpbp |E|2p = NM NM Lk c ,l d ...Lk c ,l d T ×T [ia]=[kc],[jb]=[ld] 1 1 1 1 p p p p

But this gives the formula in the statement, and we are done. The above result is quite bad news. Indeed, we cannot reconstruct the glow of H ⊗ K from that of H,K, because the indices “get mixed”.

Let us develop now some moment machinery. Let P (p) be the set of partitions of {1, . . . , p}, with its standard order relation ≤, which is such that uu ... ≤ π ≤ | | ... |, for any π ∈ P (p). We denote by µ(π, σ) the associated M¨obius function, given by:  1 if π = σ  P µ(π, σ) = − π≤τ<σ µ(π, τ) if π < σ 0 if π 6≤ σ HADAMARD MATRICES 123

p p p! For π ∈ P (p) we set = = , where b1, . . . , b|π| are the block lenghts. π b1...b|π| b1!...b|π|! N Finally, we use the following notation, where H1,...,HN ∈ T are the rows of H: O Y Hπ(i) = Hir β∈π r∈β

With these notations, we have the following result:

Theorem 8.9. The glow moments of a matrix H ∈ MN (T) are given by

Z X |E|2p = K(π)N |π|I(π) N N T ×T π∈P (p)

P p 1 P where K(π) = σ∈P (p) µ(π, σ) σ and I(π) = N |π| [i]=[j] < Hπ(i),Hπ(j) >. Proof. We know from Proposition 8.7 that the moments are given by: Z X Hi1x1 ...Hipxp |E|2p = N N Hj y ...Hj y T ×T [i]=[j],[x]=[y] 1 1 p p

With σ = ker x, ρ = ker y, we deduce that the moments of |E|2 decompose over partitions, R |E|2p = R P C(σ, ρ), with the contributions being as follows: TN ×TN TN σ,ρ∈P (p)

X X Hi1x1 ...Hipxp ai1 . . . aip C(σ, ρ) = δ[x],[y] · Hj y ...Hj y aj . . . aj ker x=σ,ker y=ρ ij 1 1 p p 1 p

We have C(σ, ρ) = 0 unless σ ∼ ρ, in the sense that σ, ρ must have the same block P structure. The point now is that the sums of type ker x=σ can be computed by using the M¨obiusinversion formula. We obtain a formula as follows: X Y C(σ, ρ) = δσ∼ρ µ(π, σ) C|β|(a) π≤σ β∈π

Here the functions on the right are by deﬁnition given by:

X X Hi x ...Hi x ai . . . ai C (a) = 1 r · 1 r r H ...H a . . . a x ij j1x jrx j1 jr X ai . . . ai = < H ...H ,H ...H > · 1 r i1 ir j1 jr a . . . a ij j1 jr 124 TEO BANICA

p Now since there are σ partitions having the same block structure as σ, we obtain: Z Z ! 2p X X X Y |Ω| = µ(π, σ) C|β|(a) N N N T ×T T π∈P (p) σ∼ρ µ≤σ β∈π   X X p Z Y =  µ(π, σ)  C|β|(a) σ N π∈P (p) σ∈P (p) T β∈π But this gives the formula in the statement, and we are done. Let us discuss now the asymptotic behavior of the glow. For this purpose, we first study the coefficients K(π) in Theorem 8.9. We have here the following result: P p Proposition 8.10. K(π) = π≤σ µ(π, σ) σ has the following properties: K(π) 0 0 (1) Ke(π) = p! is multiplicative: Ke(ππ ) = Ke(π)Ke(π ). P |σ|−1 p (2) K(uu ... u) = σ∈P (p)(−1) (|σ| − 1)! σ . Pp r−1 P p 2 (3) K(uu ... u) = (−1) (r − 1)!Cpr, where Cpr = . r=1 p=a1+...+ar a1,...,ar Proof. (1) We use the fact that µ(ππ0, σσ0) = µ(π, σ)µ(π0, σ0), which is a well-known property of the Möbiusfunction, which can be proved by recurrence. Now if b1, . . . , bs 0 and c1, . . . , ct are the block lengths of σ, σ , we obtain, as claimed: X 1 1 Ke(ππ0) = µ(ππ0, σσ0) · · b1! . . . bs! c1! . . . ct! ππ0≤σσ0 X 1 1 = µ(π, σ)µ(π0, σ0) · · b1! . . . bs! c1! . . . ct! π≤σ,π0≤σ0 = Ke(π)Ke(π0) (2) We use here the formula µ(uu ... u, σ) = (−1)|σ|−1(|σ| − 1)!, which once again is well-known, and can be proved by recurrence on |σ|. We obtain, as claimed: X p X p K(uu ... u) = µ(uu ... u, σ) = (−1)|σ|−1(|σ| − 1)! σ σ σ∈P (p) σ∈P (p) (3) By using the formula in (2), and summing over r = |σ|, we obtain: p X X p K(uu ... u) = (−1)r−1(r − 1)! σ r=1 |σ|=r p p Now if we denote by a1, . . . , ar with ai ≥ 1 the block lengths of σ, then = . σ a1,...,ar p On the other hand, given a1, . . . , ar ≥ 1 with a1 + ... + ar = p, there are exactly a1,...,ar partitions σ having these numbers as block lengths, and this gives the result. HADAMARD MATRICES 125

Now let us take a closer look at the integrals I(π). We have here: Proposition 8.11. Consider the one-block partition uu ... u ∈ P (p). (1) I(uu ... u) = #{i, j ∈ {1,...,N}p|[i] = [j]}. R P 2p (2) I(uu ... u) = N | ai| da. T i P p N! (3) I(uu ... u) = σ∈P (p) σ (N−|σ|)! . Pp−1 N! P p 2 (4) I(uu ... u) = Cpr , where Cpr = . r=1 (N−r)! p=b1+...+br b1,...,br Proof. (1) This follows indeed from the following computation: X 1 X I(uu ... u) = < H ...H ,H ...H >= 1 N i1 ir j1 jr [i]=[j] [i]=[j] (2) This follows from the following computation: Z 2p Z X X ai1 . . . aip n o ai = da = # i, j [i] = [j] N N a . . . a T i T ij j1 jp (3) If we let σ = ker i in the above formula of I(uu ... u), we obtain: n o X I(uu ... u) = # i, j ker i = σ, [i] = [j] σ∈P (p) N! p Now since there are (N−|σ|)! choices for i, and then σ for j, this gives the result. (4) If we set r = |σ|, the formula in (3) becomes: p−1 X N! X p I(uu ... u) = (N − r)! σ r=1 σ∈P (p),|σ|=r p Now since there are exactly permutations σ ∈ P (p) having b1, . . . , br as block b1,...,br lengths, the sum on the right equals P p 2, as claimed. p=b1+...+br b1,...,br In general, the integrals I(π) can be estimated as follows:

Proposition 8.12. Let H ∈ MN (T), having its rows pairwise orthogonal. (1) I(| | ... |) = N p. (2) I(| | ... | π) = N aI(π), for any π ∈ P (p − a). p (3) |I(π)| . p!N , for any π ∈ P (p). Proof. (1) Since the rows of H are pairwise orthogonal, we have: p X Y X X p I(| | ... |) = δir,jr = δij = 1 = N [i]=[j] r=1 [i]=[j] i (2) This follows by the same computation as the above one for (1). 126 TEO BANICA

(3) We have indeed the following estimate: n o X Y X p |I(π)| ≤ 1 = 1 = # i, j ∈ {1,...,N} [i] = [j] ' p!N [i]=[j] β∈π [i]=[j] Thus we have obtained the formula in the statement, and we are done. We have now all needed ingredients for a universality result:

Theorem 8.13. The glow of a complex Hadamard matrix H ∈ MN (T) is given by: 1 Z |E|2p p = 1 − N −1 + O(N −2) p! TN ×TN N 2 In particular, E/N becomes complex Gaussian in the N → ∞ limit. Proof. We use the moment formula in Theorem 8.9. By using Proposition 8.12 (3), we conclude that only the p-block and (p − 1)-block partitions contribute at order 2, so: Z p |E|2p = K(| | ... |)N pI(| | ... |) + K(u| ... |)N p−1I(u| ... |) + O(N 2p−2) TN ×TN 2 Now by dividing by N 2p and then by using the various formulae in Proposition 8.10, Proposition 8.11 and Proposition 8.12 above, we obtain, as claimed: Z 2p |E| p p! 2N − 1 −2 = p! − · 2 + O(N ) TN ×TN N 2 2 N Finally, since the law of E is invariant under centered rotations in the complex plane, this moment formula gives as well the last assertion.

Let us study now the glow of the Fourier matrices, F = FG. We use the standard P formulae FixFiy = Fi,x+y, F ix = Fi,−x and x Fix = Nδi0. We ﬁrst have:

Proposition 8.14. For a Fourier matrix FG we have ( )

X X I(π) = # i, j [i] = [j], ir = jr, ∀β ∈ π r∈β r∈β with all the indices, and with the sums at right, taken inside G. Proof. The basic components of the integrals I(π) are given by: * + 1 Y Y 1 D E Fi , Fj = FP i ,FP i = δP i ,P j N r r N r∈β r r∈β r r∈β r r∈β r r∈β r∈β But this gives the formula in the statement, and we are done. We have the following interpretation of the above integrals: HADAMARD MATRICES 127

Proposition 8.15. For any partition π we have the formula Z ! Y 1 X 2|β| I(π) = 2 |Hij| da N N T b∈π ij ∗ where H = F AF , with F = FG and A = diag(a0, . . . , aN−1). Proof. We have the following computation:

X FiyFjx ai H = F ∗AF =⇒ |H |2 = · xy F F a ij ix jy j

X Fj1x ...Fjpx Fi1y ...Fipy ai1 . . . aip =⇒ |H |2p = · · xy F ...F F ...F a . . . a ij i1x ipx j1y jpy j1 jp

X 2p X 2 ai1 . . . aip =⇒ |Hxy| = < Hi1 ...Hip ,Hj1 ...Hjp > · a . . . a xy ij j1 jp But this gives the formula in the statement, and we are done. Regarding now the glow estimates, we ﬁrst have the following result:

Proposition 8.16. For FG we have the estimate p p−1 I(π) = b1! . . . b|π|!N + O(N ) where b1, . . . , b|π| with b1 + ... + b|π| = p are the block lengths of π. Proof. With σ = ker i we obtain: ( )

X X X I(π) = # i, j ker i = σ, [i] = [j], ir = jr, ∀β ∈ π σ∈P (p) r∈β r∈β

N! |σ| Since there are (N−|σ|)! ' N choices for i satisfying ker i = σ, and then there are p σ = O(1) choices for j satisfying [i] = [j], we conclude that the main contribution comes from σ = | | ... |, and so we have: ( )

X X p−1 I(π) = # i, j ker i = | | ... |, [i] = [j], ir = jr, ∀β ∈ π + O(N ) r∈β r∈β Now the condition ker i = | | ... | tells us that i must have distinct entries, and there N! p are (N−p)! ' N choices for such multi-indices i. Regarding now the indices j, the main contribution comes from those obtained from i by permuting the entries over the blocks of π, and since there are b1! . . . b|π|! choices here, this gives the result. At the second order now, the estimate is as follows: 128 TEO BANICA

Proposition 8.17. For FG we have the formula ! I(π) X X bi bj 1 X bi = 1 + − N −1 + O(N −2) b ! . . . b !N p c c 2 2 1 s i

i1 . . . ic k1 . . . ka−c j1 . . . jc l1 . . . la−c j1 . . . jc k1 . . . ka−c i1 . . . ic l1 . . . la−c | {z } | {z } | {z } | {z } c a−c c b−c Let us compute now I1(a, b). We cannot have c = 0, 1, and once c ≥ 2 is given, we have a b p−1 p−2 c , c choices for the positions of the i, j variables in the upper row, then N +O(N ) choices for the variables in the upper row, and then finally we have a!b! permutations which can produce the lower row. We therefore obtain: X ab I1(a, b) = a!b! N p−1 + O(N p−2) c c c≥2 In the general case now, a similar discussion applies. Indeed, the constraint of type P P r∈β ir = r∈β jr with [ir|r ∈ β] 6= [jr|r ∈ β] cannot affect ≤ 1 blocks, because we are not in the non-arithmetic case, and cannot affect either ≥ 3 blocks, because affecting ≥ 3 HADAMARD MATRICES 129 blocks would require ≥ 2 constraints. Thus this condition affects exactly 2 blocks, and if we let i < j be the indices in {1, . . . , s} corresponding to these 2 blocks, we obtain: X X bi bj I1(π) = b ! . . . b ! N p−1 + O(N p−2) 1 s c c i

Proposition 8.18. For FG with G = ZN1 × ... × ZNk we have the formula I(uu) = N(4N 3 − 11N + 2e + 7) where e ∈ {0, 1, . . . , k} is the number of even numbers among N1,...,Nk. P P Proof. We use the fact that, when dealing with the conditions r∈β ir = r∈β jr deﬁning the quantities I(π), one can always erase some of the variables ir, jr, as to reduce to the “purely arithmetic” case, {ir|r ∈ β} ∩ {jr|r ∈ β} = ∅. We have: I(uu) = I◦(uu) + Iari(uu) Let us compute now Iari(uu). There are 3 contributions to this quantity, namely: iijj (1) Case (jjii), with i 6= j, 2i = 2j. Since 2(i1, . . . , ik) = 2(j1, . . . , jk) corresponds to the collection of conditions 2ir = 2jr, inside ZNr , which each have 1 or 2 solutions, depending on whether Nr is odd or even, the contribution here is: ari I1 (uu) = #{i 6= j|2i = 2j} = #{i, j|2i = 2j} − #{i, j|i = j} = 2eN − N = (2e − 1)N 130 TEO BANICA

iijk (2) Case (jkii), with i, j, k distinct, 2i = j + k. The contribution here is: ari I2 (uu) = 4#{i, j, k distinct|2i = j + k} = 4#{i 6= j|2i − j 6= i, j} = 4#{i 6= j|2i 6= 2j} = 4(#{i, j|i 6= j} − #{i 6= j|2i = 2j}) = 4(N(N − 1) − (2e − 1)N) = 4N(N − 2e) ijkl (3) Case (klij), with i, j, k, l distinct, i + j = k + l. The contribution here is: ari I3 (uu) = 4#{i, j, k, l distinct|i + j = k + l} = 4#{i, j, k distinct|i + j − k 6= i, j, k} = 4#{i, j, k distinct|i + j − k 6= k} = 4#{i, j, k distinct|i 6= 2k − j} We can split this quantity over two cases, 2j 6= 2k and 2j = 2k, and we obtain: ari I3 (uu) = 4(#{i, j, k distinct|2j 6= 2k, i 6= 2k − j} +#{i, j, k distinct|2j = 2k, i 6= 2k − j}) The point now is that in the ﬁrst case, 2j 6= 2k, the numbers j, k, 2k − j are distinct, while in the second case, 2j = 2k, we simply have 2k − j = j. Thus, we obtain: ! ari X X I3 (uu) = 4 #{i|i 6= j, k, 2k − j} + #{i|i 6= j, k} j6=k,2j6=2k j6=k,2j=2k = 4(N(N − 2e)(N − 3) + N(2e − 1)(N − 2)) = 4N(N(N − 3) − 2e(N − 3) + 2e(N − 2) − (N − 2)) = 4N(N 2 − 4N + 2e + 2) We can now compute the arithmetic part. This is given by: Iari(uu) = (2e − 1)N + 4N(N − 2e) + 4N(N 2 − 4N + 2e + 2) = N(2e − 1 + 4(N − 2e) + 4(N 2 − 4N + 2e + 2)) = N(4N 2 − 12N + 2e + 7) Thus the integral to be computed is given by: I(uu) = N 2(2N − 1)2 + N(4N 2 − 12N + 2e + 7) = N(4N 3 − 4N 2 + N + 4N 2 − 12N + 2e + 7) = N(4N 3 − 11N + 2e + 7) Thus we have reached to the formula in the statement, and we are done. HADAMARD MATRICES 131

We have the following asymptotic result:

Theorem 8.19. The glow of FG, with |G| = N, is given by Z 2p 1 |E| −1 −2 −3 −4 = 1 − K1N + K2N − K3N + O(N ) p! TN ×TN N p p 3p2+p−8 p p3+4p2+p−18 with K1 = 2 , K2 = 2 12 , K3 = 3 8 . K(π) I(π) Proof. We use the quantities Ke(π) = p! , Ie(π) = N p , which are such that Ke(π| ... |) = p Ke(π), Ie(π| ... |) = Ie(π). In terms of J(σ) = σ Ke(σ)Ie(σ), we have: 1 Z |E|2p = J(∅) p! TN ×TN + N −1J(u) + N −2 (J(uu) + J(uu)) + N −3 (J(uuu) + J(uu u) + J(u u u)) + O(N −4)

1 1 1 3 2 We have Ke0 = Ke1 = 1, Ke2 = 2 − 1 = − 2 , Ke3 = 6 − 2 + 2 = 3 and: 1 4 3 12 11 Ke4 = − − + − 6 = − 24 6 4 2 8

Regarding now the numbers Cpr in Proposition 8.16, these are given by: 12p p!p C = 1,C = − 1,...... ,C = ,C = p! p1 p2 2 p p,p−1 2 2 pp We deduce that I(|) = N, I(u) = N(2N − 1), I(uu) = N(6N 2 − 9N + 4) and: I(uuu) = N(24N 3 − 72N 2 + 82N − 33) By using Proposition 8.17 and Proposition 8.18, we obtain the following formula: 1 Z 1p 2p |E|2p = 1 − (2N −1 − N −2) + (6N −2 − 9N −3) p! TN ×TN 2 2 3 3 p p p + 3 N −2 − 33 N −3 − 40 N −3 4 4 5 p − 15 N −3 + O(N −4) 6

But this gives the formulae of K1,K2,K3 in the statement, and we are done. It is possible to compute the next term as well, the result being: 132 TEO BANICA

Theorem 8.20. Let G = ZN1 ×...×ZNk be a ﬁnite abelian group, and set N = N1 ...Nk. Then the glow of the associated Fourier matrix FG is given by Z 2p 1 |E| −1 −2 −3 −4 −5 = 1 − K1N + K2N − K3N + K4N + O(N ) p! TN ×TN N where the quantities K1,K2,K3,K4 are given by p K = 1 2 p3p2 + p − 8 K = 2 2 12 pp3 + 4p2 + p − 18 K = 3 3 8 8p 3 2e p p 2915p p p K = + 121 + + 416 + + 40 + 105 4 3 3 4 N 4 5 2 6 7 8 where e ∈ {0, 1, . . . , k} is the number of even numbers among N1,...,Nk. Proof. This is something that we already know, up to order 3, and the next coeﬃcient K4 can be computed in a similar way, based on results that we already have. Skipping the technical details here, we obtain the formula for K4 in the statement. The passage from Theorem 8.19 to Theorem 8.20 is quite interesting, because it shows that the glow of the Fourier matrices FG is not polynomial in N = |G|. When restricting the attention to the usual Fourier matrices FN , the glow up to order 4 is polynomial both in N odd, and in N even, but it is not clear what happens at higher order. An interesting question here is that of computing the glow of the Walsh matrices. For n n such a matrix WN , with N = 2 , the underlying group is G = Z2 , and the numbers

n P o n CI (J1,...,Jr) = # (ai)i∈I ∈ G distinct aj = 0, ∀s are polynomial in N = 2 . j∈Js This suggests that the integrals I(π), and hence the glow, should be polynomial in N. There are many interesting questions in relation with the glow. As already mentioned in the beginning of this section, a motivation for all this comes from [57]. Also, we have as well the question of connecting the various invariants of the Hadamard matrices, such as the defect, or the algebraic invariants from sections 10-12 below, to the glow. HADAMARD MATRICES 133

9. Norm maximizers

We discuss here some further analytic questions.√ We know from Theorem 7.1 above that the Hadamard manifold XN = MN (T)∩ NUN can be analytically computed in two possible ways. The ﬁrst method is via the Hadamard determinant bound: n o N/2 XN = H ∈ MN (T) | det(H)| = N This method, going back to Hadamard’s 1893 paper [53] is of course something well- known, and we will not further comment on this. We will focus instead on the second possible method, which is something more recent [14], via the 1-norm bound: n √ o 2 XN = H ∈ NUN ||H||1 = N

This formula suggests a systematic study of the 1-norm on UN , with results about critical points, and local maximizers. In addition, in connection with the real Hadamard matrix problematics, we would like to study as well the 1-norm on ON , where the critical points and local maximizers might be diﬀerent from the real ones over UN .

We have already met such questions, in sections 1,6,7 above, in various technical con- texts. Let us begin with a short summary of the results that we already have: √ Theorem 9.1. Let U ∈ UN , and set H = NU. 2/p−1/2 (1) For p ∈ [1, 2) we have ||U||p ≤ N , with equality when H is Hadamard. 2/p−1/2 (2) For p ∈ (2, ∞] we have ||U||p ≥ N , with equality when H is Hadamard. Proof. As explained in section 6 above, all this follows from the Hölderinequality, with the remark that at p = 1, 4 we just need the Cauchy-Schwarz inequality. We have chosen here to talk about p-norms instead of just the 1-norm, and this, for two reasons. First, we have seen in section 6 that for certain technical questions regarding the circulant matrices, the natural norm to be used is in fact the 4-norm. And second, we will formulate in what follows certain conjectures regarding the 1-norm, and these conjectures are probably easier to investigate first in the p-norm setting, with p arbitrary. The above result suggests the following definition: √ Definition 9.2. Given U ∈ UN , the matrix H = NU is called:

(1) Almost Hadamard, if U locally maximizes the 1-norm on UN . (2) p-almost Hadamard, with p < 2, if U locally maximizes the p-norm on UN . (3) p-almost Hadamard, with p > 2, if U locally minimizes the p-norm on UN . (4) Absolute almost Hadamard, if it is p-almost Hadamard at any p 6= 2.

We have as well real versions of these notions, with UN replaced by ON . 134 TEO BANICA

All this might seem a bit complicated, but this is the best way of presenting things. We are mainly interested in (1), but as mentioned above, the exponent p = 4 from (3) is interesting as well, and once we have (3) we must formulate (2) as well, and ﬁnally (4) is a useful thing too, because the absolute case is sometimes easier to study. As for the “doubling” of all these notions, via the last sentence, this is necessary too, because given a function F : UN → R, an element U ∈ ON can be a local extremum of the restriction F|ON : ON → R, but not of the function F itself. And, we will see in what follows that this is the case, and in a quite surprising way, with the p-norms.

Let us ﬁrst study the critical points. Things are quite tricky here, and complete results are available so far only at p = 1. In the real case, following [14], we have:

Theorem 9.3. If U ∈ ON locally maximizes the 1-norm, then Uij 6= 0 for any i, j. Proof. Assume that U has a 0 entry. By permuting the rows we can assume that this 0 entry is in the ﬁrst row, having under it a nonzero entry in the second row. We denote by U1,...,UN the rows of U. By permuting the columns we can assume that we have a block decomposition of the following type: U 0 0 YAB 1 = U2 0 X 0 CD Here X, Y, A, B, C, D are certain vectors with nonzero entries, with A, B, C, D chosen such that each entry of A has the same sign as the corresponding entry of C, and each entry of B has sign opposite to the sign of the corresponding entry of D. Our above assumption states that X is not the null vector. For t > 0 small consider the matrix U t obtained by rotating by t the ﬁrst two rows of U. In row notation, this matrix is given by:       cos t sin t U1 cos t · U1 + sin t · U2 − sin t cos t    U2  − sin t · U1 + cos t · U2 t  1      U =    U3  =  U3   .       ..  ...  ...  1 UN UN We make the convention that the lower-case letters denote the 1-norms of the corresponding upper-case vectors. According to the above sign conventions, we have: N t X ||U ||1 = || cos t · U1 + sin t · U2||1 + || − sin t · U1 + cos t · U2||1 + ui i=3 N X = (cos t + sin t)(x + y + b + c) + (cos t − sin t)(a + d) + ui i=3

= ||U||1 + (cos t + sin t − 1)(x + y + b + c) + (cos t − sin t − 1)(a + d) HADAMARD MATRICES 135

By using sin t = t + O(t2) and cos t = 1 + O(t2) we obtain:

t 2 ||U ||1 = ||U||1 + t(x + y + b + c) − t(a + d) + O(t ) 2 = ||U||1 + t(x + y + b + c − a − d) + O(t ) In order to conclude, we have to prove that U cannot be a local maximizer of the 1-norm. This will basically follow by comparing the norm of U to the norm of U t, with t > 0 small or t < 0 big. However, since in the above computation it was technically convenient to assume t > 0, we actually have three cases: Case 1: b + c > a + d. Here for t > 0 small enough the above formula shows that we t have ||U ||1 > ||U||1, and we are done. Case 2: b + c = a + d. Here we use the fact that X is not null, which gives x > 0. Once t again for t > 0 small enough we have ||U ||1 > ||U||1, and we are done. Case 3: b + c < a + d. In this case we can interchange the ﬁrst two rows of U and restart the whole procedure: we fall in Case 1, and we are done again. In the complex case, following [17], we have a similar result:

Theorem 9.4. If U ∈ UN locally maximizes the 1-norm, then Uij 6= 0 for any i, j. Proof. We use the same method as in the real case, namely a “rotation trick”. Let us denote by U1,...,UN the rows of U, and let us perform a rotation of U1,U2:

t U1 cos t · U1 − sin t · U2 t = U2 sin t · U1 + cos t · U2 In order to compute the 1-norm, let us permute the columns of U, in such a way that the ﬁrst two rows look as follows, with X, Y, A, B having nonzero entries: U 0 0 YA 1 = U2 0 X 0 B The rotated matrix will look then as follows: t U1 0 − sin t · X cos t · Y cos t · A − sin t · B t = U2 0 cos t · X sin t · y sin t · A + cos t · B Our claim is that X,Y must be empty. Indeed, if A and B are not empty, let us ﬁx a column index k for both A, B, and set α = Ak, β = Bk. We have then:

t t |(U1)k| + |(U2)k| = | cos t · α − sin t · β| + | sin t · α + cos t · β| q = cos2 t · |α|2 + sin2 t · |β|2 − sin t cos t(αβ¯ + βα¯) q + sin2 t · |α|2 + cos2 t · |β|2 + sin t cos t(αβ¯ + βα¯) 136 TEO BANICA

Since α, β 6= 0, the above function is derivable at t = 0, and we obtain: ∂ (|(U t) | + |(U t) |) sin 2t(|β|2 − |α|2) − cos 2t(αβ¯ + βα¯) 1 k 2 k = ∂t 2pcos2 t · |α|2 + sin2 t · |β|2 − sin t cos t(αβ¯ + βα¯) sin 2t(|α|2 − |β|2) + cos 2t(αβ¯ + βα¯) + 2psin2 t · |α|2 + cos2 t · |β|2 + sin t cos t(αβ¯ + βα¯) Thus at t = 0, we obtain the following formula: ∂ (|(U t) | + |(U t) |) αβ¯ + βα¯ 1 1 1 k 2 k (0) = − ∂t 2 |β| |α| t Now since U locally maximizes the 1-norm, both directional derivatives of ||U ||1 must be negative in the limit t → 0. On the other hand, if we denote by C the contribution coming from the right (which might be zero in the case where A and B are empty), i.e. the sum over k of the above quantities, we have: t ∂||U ||1 ∂ = (| cos t| + | sin t|)(||X||1 + ||Y ||1) + C ∂t t=0+ ∂t t=0+

= (− sin t + cos t) (||X||1 + ||Y ||1) + C t=0

= ||X||1 + ||Y ||1 + C As for the derivative at left, this is given by the following formula: t ∂||U ||1 ∂ = (| cos t| + | sin t|)(||X||1 + ||Y ||1) + C ∂t t=0− ∂t t=0−

= (− sin t − cos t) (||X||1 + ||Y ||1) + C t=0

= −||X||1 − ||Y ||1 + C We therefore obtain the following inequalities, where C is as above:

||X||1 + ||Y ||1 + C ≤ 0

−||X||1 − ||Y ||1 + C ≤ 0

Consider now the matrix obtained from U by interchanging U1,U2. Since this matrix must be as well a local maximizer of the 1-norm, and since the above formula shows that C changes its sign when interchanging U1,U2, we obtain:

||X||1 + ||Y ||1 − C ≤ 0

−||X||1 − ||Y ||1 − C ≤ 0

The four inequalities that we have give altogether ||X||1 + ||Y ||1 = C = 0, and from ||X||1 + ||Y ||1 = 0 we obtain that both X,Y must be empty, as claimed. HADAMARD MATRICES 137

As a conclusion, up to a permutation of the columns, the ﬁrst two rows must be of the following form, with A, B having only nonzero entries: U 0 A 1 = U2 0 B

By permuting the rows of U, the same must hold for any two rows Ui,Uj. Now since U cannot have a zero column, we conclude that U cannot have zero entries, as claimed. Let us compute now the critical points. Following [17], we have:

∗ Theorem 9.5. Let ϕ : [0, ∞) → R be a diﬀerentiable function. A matrix U ∈ UN is a critical point of the quantity X F (U) = ϕ(|Uij|) ij ∗ 0 precisely when WU is self-adjoint, where Wij = sgn(Uij)ϕ (|Uij|). ¯ Proof. We regard UN as a real algebraic manifold, with coordinates Uij, Uij. This manifold consists by deﬁnition of the zeroes of the following polynomials: X ¯ Aij = UikUjk − δij k

Since UN is smooth, and so is a diﬀerential manifold in the usual sense, it follows from the general theory of Lagrange multipliers that a given matrix U ∈ UN is a critical point of F precisely when the condition dF ∈ span(dAij) is satisﬁed. Regarding the space span(dAij), this consists of the following quantities: X X ¯ ¯ MijdAij = Mij(UikdUjk + UjkdUik) ij ijk X t ¯ X ¯ = (M U)jkdUjk + (MU)ikdUik jk ik X t ¯ X ¯ = (M U)ijdUij + (MU)ijdUij ij ij

In order to compute dF , observe ﬁrst that, with Sij = sgn(Uij), we have: q ¯ ¯ ¯ UijdUij + UijdUij 1 ¯ ¯ d|Uij| = d UijUij = = (SijdUij + SijdUij) 2|Uij| 2 0 We therefore obtain, with Wij = sgn(Uij)ϕ (|Uij|) as in the statement: X X 1 X dF = d (ϕ(|U |)) = ϕ0(|U |)d|U | = W dU¯ + W¯ dU ij ij ij 2 ij ij ij ij ij ij ij 138 TEO BANICA

We conclude that U ∈ UN is a critical point of F if and only if there exists a matrix M ∈ MN (C) such that the following two conditions are satisﬁed: W = 2M tU, W¯ = 2MU¯ Now observe that these two equations can be written as follows: 1 1 M t = WU ∗ ,M t = UW ∗ 2 2 ∗ ∗ Summing up, the critical point condition on U ∈ UN simply reads WU = UW , which ∗ means that the matrix WU must be self-adjoint, as claimed. In order to process the above result, we can use the following notion: P Deﬁnition 9.6. Given U ∈ UN , we consider its “color decomposition” U = r>0 rUr, with Ur ∈ MN (T ∪ {0}) containing the phase components at r > 0, and we call U: ∗ ∗ (1) Semi-balanced, if UrU and U Ur, with r > 0, are all self-adjoint. ∗ ∗ (2) Balanced, if UrUs and Ur Us, with r, s > 0, are all self-adjoint.

These conditions are quite natural, because for a unitary matrix U ∈ UN , the relations UU ∗ = U ∗U = 1 translate as follows, in terms of the color decomposition: X ∗ X ∗ rUrU = rU Ur = 1 r>0 r>0

X ∗ X ∗ rsUrUs = rsUr Us = 1 r,s>0 r,s>0 Thus, our balancing conditions express the fact that the various components of the above sums all self-adjoint. Now back to our critical point questions, we have:

∗ Theorem 9.7. For a matrix U ∈ UN , the following are equivalent: P (1) U is a critical point of F (U) = ij ϕ(|Uij|), for any ϕ : [0, ∞) → R. (2) U is a critical point of all the p-norms, with p ∈ [1, ∞). (3) U is semi-balanced, in the above sense. Proof. We use Theorem 9.5 above. The matrix constructed there is given by: ∗ X 0 ¯ (WU )ij = sgn(Uik)ϕ (|Uik|)Ujk k X 0 X ¯ = ϕ (r) sgn(Uik)Ujk

r>0 k,|Uik|=r X 0 X ¯ = ϕ (r) (Ur)ikUjk r>0 k X 0 ∗ = ϕ (r)(UrU )ij r>0 HADAMARD MATRICES 139

∗ P 0 ∗ Thus we have WU = r>0 ϕ (r)UrU , and when ϕ : [0, ∞) → R varies, either as an arbitrary diﬀerentiable function, or as a power function ϕ(x) = xp wirh p ∈ [1, ∞), the individual components of this sum must be all self-adjoint, as desired. In practice now, most of the known examples of semi-balanced matrices are actually balanced. We have the following collection of simple facts, regarding such matrices: Proposition 9.8. The class of balanced matrices is as follows: √ (1) It contains the matrices U = H/ N, with H ∈ MN (C) Hadamard. (2) It is stable under transposition, complex conjugation, and taking adjoints. (3) It is stable under taking tensor products. (4) It is stable under the Hadamard equivalence relation. 1 (5) It contains the matrix VN = N (2IN − N1N ), where IN is the all-1 matrix. Proof. All these results are elementary, the proof being as follows: (1) Here U ∈ UN follows from the Hadamard condition, and since there is only one √ color component, namely U1/ N = H, the balancing condition is satisﬁed as well. P (2) Assuming that U = r>0 rUr is a color decomposition of a given matrix U ∈ UN , the following are color decompositions too, and this gives the assertions: t X t ¯ X ¯ ∗ X ∗ U = rUr , U = rUr ,U = rUr r>0 r>0 r>0 P P (3) Assuming that U = r>0 rUr and V = s>0 sVs are the color decompositions of two given unitary matrices U, V , we have: X X X U ⊗ V = rs · Ur ⊗ Vs = p Ur ⊗ Vs r,s>0 p>0 p=rs P Thus the color components of W = U ⊗ V are the matrices Wp = p=rs Ur ⊗ Vs, and it follows that if U, V are both balanced, then so is W = U ⊗ V . (4) We recall that the Hadamard equivalence consists in permuting rows and columns, and switching signs on rows and columns. Since all these operations correspond to certain ∗ ∗ conjugations at the level of the matrices UrUs ,Ur Us, we obtain the result. (5) The matrix in the statement, which goes back to [20], is as follows: 2 − N 2 ... 2  1  2 2 − N... 2  VN =   N  ......  2 2 ... 2 − N Observe that this matrix is indeed unitary, its rows being of norm one, and pairwise orthogonal. The color components of this matrix being V2/N−1 = 1N and V2/N = IN − 1N , it follows that this matrix is balanced as well, as claimed. 140 TEO BANICA

Let us look now more in detail at VN , and at the matrices having similar properties. Following [20], let us call (a, b, c) pattern any matrix M ∈ MN (0, 1), with N = a + 2b + c, such that any two rows look as follows, up to a permutation of the columns: 0 ... 0 0 ... 0 1 ... 1 1 ... 1 0 ... 0 1 ... 1 0 ... 0 1 ... 1 | {z } | {z } | {z } | {z } a b b c As explained in [20], there are many interesting examples of (a, b, c) patterns, coming from the balanced incomplete block designs (BIBD), and all these examples can produce two-entry unitary matrices, by replacing the 0, 1 entries with suitable numbers x, y. Now back to the matrix VN from Proposition 9.8 (5), observe that this matrix comes from a (0, 1,N − 2) pattern. And also, independently of this, this matrix has the remark- able property of being at the same time circulant and self-adjoint. We have in fact the following result, generalizing Proposition 9.8 (5): Theorem 9.9. The following matrices are balanced: (1) The orthogonal matrices coming from (a, b, c) patterns. (2) The unitary matrices which are circulant and self-adjoint. Proof. These observations basically go back to [20], the proofs being as follows: (1) If we denote by P,Q ∈ MN (0, 1) the matrices describing the positions of the 0, 1 entries inside the pattern, then we have the following formulae:

t t PP = P P = aIN + b1N t t QQ = Q Q = cIN + b1N t t t t PQ = P Q = QP = Q P = bIN − b1N Since all these matrices are symmetric, U is balanced, as claimed. (2) Assume that U ∈ UN is circulant, Uij = γj−i, and in addition self-adjoint, which meansγ ¯i = γ−i. Consider the following sets, which must satisfy Dr = −Dr:

Dr = {k : |γr| = k} In terms of these sets, we have the following formula:

∗ X ¯ (UrUs )ij = (Ur)ik(Us)jk k X = δ|γk−i|,r sgn(γk−i) · δ|γk−j |,s sgn(¯γk−j) k X = sgn(γk−i) sgn(¯γk−j)

k∈(Dr+i)∩(Ds+j) HADAMARD MATRICES 141

With k = i + j − m we obtain, by using Dr = −Dr, and thenγ ¯i = γ−i:

∗ X (UrUs )ij = sgn(γj−m) sgn(¯γi−m)

m∈(−Dr+j)∩(−Ds+i) X = sgn(γj−m) sgn(¯γi−m)

m∈(Dr+i)∩(Dr+j) X = sgn(¯γm−j) sgn(γm−i)

m∈(Dr+i)∩(Dr+j) Now by interchanging i ↔ j, and with m → k, this formula becomes:

∗ X (UrUs )ji = sgn(¯γk−i) sgn(γk−j)

k∈(Dr+i)∩(Dr+j)

∗ We recognize here the complex conjugate of (UrUs )ij, as previously computed above, ∗ ∗ and we therefore deduce that UrUs is self-adjoint. The proof for Ur Us is similar. Summarizing, the study of the critical points alone leads to some interesting combinatorics. There are of course many questions regarding the balanced and semi-balanced matrices, for instance in connection with design theory. See [17].

Let us compute now derivatives. As in Theorem 9.5, it is convenient to do the computations in a more general framework, where we have a function as follows:

X 2 F (U) = ψ(|Uij| ) ij In order to study the local extrema of these quantities, consider the following function, depending on t > 0 small:

tA X tA 2 f(t) = F (Ue ) = ψ(|(Ue )ij| ) ij

Here U ∈ UN is an arbitrary unitary matrix, and A ∈ MN (C) is assumed to be anti- ∗ A hermitian, A = −A, with this latter assumption needed for having e ∈ UN . Let us ﬁrst compute the derivative of f. We have: Proposition 9.10. We have the following formula, h i 0 X 0 tA 2 tA tA f (t) = 2 ψ (|(Ue )ij| )Re (UAe )ij(Ue )ij ij valid for any U ∈ UN , and any A ∈ MN (C) anti-hermitian. 142 TEO BANICA

Proof. The matrices U, etA being both unitary, we have:

tA 2 tA tA |(Ue )ij| = (Ue )ij(Ue )ij tA tA ∗ = (Ue )ij((Ue ) )ji tA tA∗ ∗ = (Ue )ij(e U )ji tA −tA ∗ = (Ue )ij(e U )ji We can now diﬀerentiate our function f, and by using once again the unitarity of the matrices U, etA, along with the formula A∗ = −A, we obtain:

0 X 0 tA 2 tA −tA ∗ tA −tA ∗ f (t) = ψ (|(Ue )ij| ) (UAe )ij(e U )ji − (Ue )ij(e AU )ji ij h i X 0 tA 2 tA −tA ∗ ∗ tA −tA ∗ ∗ = ψ (|(Ue )ij| ) (UAe )ij((e U ) )ij − (Ue )ij((e AU ) )ij ij h i X 0 tA 2 tA tA tA tA = ψ (|(Ue )ij| ) (UAe )ij(Ue )ij + (Ue )ij(UAe )ij ij But this gives the formula in the statement, and we are done. Before computing the second derivative, let us evaluate f 0(0). In terms of the color P decomposition U = r>0 rUr of our matrix, the result is as follows: Proposition 9.11. We have the following formula,

0 X 0 2 ∗ f (0) = 2 rψ (r )Re [T r(Ur UA)] r>0 where Ur ∈ MN (T ∪ {0}) are the color components of U. Proof. We use the formula in Proposition 9.10 above. At t = 0, we obtain:

0 X 0 2 f (0) = 2 ψ (|Uij| )Re (UA)ijU ij ij Consider now the color decomposition of U. We have the following formulae:

X 2 X 2 Uij = r(Ur)ij =⇒ |Uij| = r |(Ur)ij| r>0 r>0 0 2 X 0 2 =⇒ ψ (|Uij| ) = ψ (r )|(Ur)ij| r>0 Now by getting back to the above formula of f 0(0), we obtain:

0 X 0 2 X f (0) = 2 ψ (r ) Re (UA)ijU ij|(Ur)ij| r>0 ij HADAMARD MATRICES 143

0 X 0 2 X h i f (0) = 2 rψ (r ) Re (UA)ij(Ur)ij r>0 ij

But this gives the formula in the statement, and we are done.

Let us compute now the second derivative. The result here is as follows:

Proposition 9.12. We have the following formula,

00 X 00 2 2 f (0) = 4 ψ (|Uij| )Re (UA)ijU ij ij X 0 2 2 +2 ψ (|Uij| )Re (UA )ijU ij ij X 0 2 2 +2 ψ (|Uij| )|(UA)ij| ij valid for any U ∈ UN , and any A ∈ MN (C) anti-hermitian. Proof. We use the formula in Proposition 9.10 above, namely: h i 0 X 0 tA 2 tA tA f (t) = 2 ψ (|(Ue )ij| )Re (UAe )ij(Ue )ij ij

Since the real part on the right, or rather its double, appears as the derivative of the tA 2 quantity |(Ue )ij| , when diﬀerentiating a second time, we obtain:

h i2 00 X 00 tA 2 tA tA f (t) = 4 ψ (|(Ue )ij| )Re (UAe )ij(Ue )ij ij h i0 X 0 tA 2 tA tA +2 ψ (|(Ue )ij| )Re (UAe )ij(Ue )ij ij

In order to compute now the missing derivative, observe that we have:

h i0 tA tA 2 tA tA tA tA (UAe )ij(Ue )ij = (UA e )ij(Ue )ij + (UAe )ij(UAe )ij

2 tA tA tA 2 = (UA e )ij(Ue )ij + |(UAe )ij| 144 TEO BANICA

Summing up, we have obtained the following formula: h i2 00 X 00 tA 2 tA tA f (t) = 4 ψ (|(Ue )ij| )Re (UAe )ij(Ue )ij ij h i X 0 tA 2 2 tA tA +2 ψ (|(Ue )ij| )Re (UA e )ij(Ue )ij ij X 0 tA 2 tA 2 +2 ψ (|(Ue )ij| )|(UAe )ij| ij But at t = 0 this gives the formula in the statement, and we are done. √ For the function ψ(x) = x, corresponding to the functional F (U) = ||U||1, there are some simpliﬁcations, that we will work out now in detail. First, we have:

∗ Proposition 9.13. Let U ∈ UN . For the function F (U) = ||U||1 we have the formula 2 X Im (UA)ijSij f 00(0) = Re T r(S∗UA2) + |U | ij ij valid for any anti-hermitian matrix A, where Uij = Sij|Uij|. √ Proof. We use the formula in Proposition 9.12 above, with ψ(x) = x. The derivatives 0 √1 00 1√ are here ψ (x) = 2 x and ψ (x) = − 4x x , and we obtain:

2 2 2 X Re (UA)ijU ij X Re (UA )ijU ij X |(UA)ij| f 00(0) = − + + |U |3 |U | |U | ij ij ij ij ij ij

∗ Theorem 9.14. A matrix U ∈ UN locally maximizes the one-norm on UN precisely when ∗ S U is self-adjoint, where Sij = sgn(Uij), and when 2 X Im (UA)ijSij T r(S∗UA2) + ≤ 0 |U | ij ij holds, for any anti-hermitian matrix A ∈ MN (C). HADAMARD MATRICES 145

Proof. According to Theorem 9.5 and Proposition 9.13, the local maximizer condition requires X = S∗U to be self-adjoint, and the following inequality to be satisﬁed: 2 X Im (UA)ijSij Re T r(S∗UA2) + ≤ 0 |U | ij ij Now observe that since both X and A2 are self-adjoint, we have: 1 Re T r(XA2) = T r(XA2) + T r(A2X) = T r(XA2) 2 Thus we can remove the real part, and we obtain the inequality in the statement. In order to further improve the above result, we will need:

Proposition 9.15. For a self-adjoint matrix X ∈ MN (C), the following are equivalent: 2 (1) T r(XA ) ≤ 0, for any anti-hermitian matrix A ∈ MN (C). 2 (2) T r(XB ) ≥ 0, for any hermitian matrix B ∈ MN (C). (3) T r(XC) ≥ 0, for any positive matrix C ∈ MN (C). (4) X ≥ 0. Proof. These equivalences are well-known, the proof being as follows: (1) =⇒ (2) follows by taking B = iA. (2) =⇒ (3) follows by taking C = B2. (3) =⇒ (4) follows by diagonalizing X, and then√ taking C to be diagonal. (4) =⇒ (1) is clear as well, because with Y = X we have: T r(XA2) = T r(Y 2A2) = T r(YA2Y ) = −T r((YA)(YA)∗) ≤ 0 Thus, the above four conditions are indeed equivalent. We would like to discuss as well the real case, and we will need here:

Proposition 9.16. For a symmetric matrix X ∈ MN (R), the following are equivalent: (1) T r(XA2) ≤ 0, for any antisymmetric matrix A. (2) The sum of the two smallest eigenvalues of X is positive. P Proof. In terms of the vector a = ij Aijei ⊗ ej, which is antisymmetric, we have: T r(XA2) =< X, A2 >= − < AX, A >= − < a, (1 ⊗ X)a > Thus the condition (1) is equivalent to P (1 ⊗ X)P being positive, with P being the orthogonal projection on the antisymmetric subspace in RN ⊗ RN . For any two eigenvectors xi ⊥ xj of X, with eigenvalues λi, λj, we have:

P (1 ⊗ X)P (xi ⊗ xj − xj ⊗ xi) = P (λjxi ⊗ xj − λixj ⊗ xi) λ + λ = i j (x ⊗ x − x ⊗ x ) 2 i j j i Thus, we obtain the conclusion in the statement. 146 TEO BANICA

Following [17], we can now formulate a ﬁnal result on the subject, as follows:

∗ Theorem 9.17. Given U ∈ UN , set Sij = sgn(Uij), and X = S U.

(1) U locally maximizes the 1-norm on UN precisely when X ≥ 0, and when 2 X Re (UB)ijSij Φ(U, B) = T r(XB2) − |U | ij ij

is positive, for any hermitian matrix B ∈ MN (C). (2) If U ∈ ON , this matrix locally maximizes the 1-norm on ON precisely when X is self-adjoint, and the sum of its two smallest eigenvalues is positive. Proof. Here (1) follows from Theorem 9.14, by setting A = iB, and by using Proposition 9.15, which shows that we must have indeed X ≥ 0. As for (2), this follows from (1), with the remark that the right term vanishes, and from Proposition 9.16. In practice now, let us ﬁrst discuss the real case. The following result, involving the notion of (a, b, c) pattern appearing in Theorem 9.9, was proved in [20]: Theorem 9.18. If U = U(x, y) is orthogonal, coming from an (a, b, c) pattern, with (N(a − b) + 2b)|x| + (N(c − b) + 2b)|y| ≥ 0 √ the matrix H = NU is almost Hadamard, in a real sense. Proof. Since any row of U consists of a + b copies of x and b + c copies of y, we have: ( (a + b)|x| + (b + c)|y| (i = j) (SU t) = ij (a − b)|x| + (c − b)|y| (i 6= j) Now observe that we can write the matrix SU t as follows: t SU = 2b(|x| + |y|)1N + ((a − b)|x| + (c − b)|y|)NJN

= 2b(|x| + |y|)(1N − JN ) + ((N(a − b) + 2b)|x| + (N(c − b) + 2b)|y|))JN t Since 1N − JN ,JN are orthogonal projections, we have SU > 0 if and only if the coeﬃcients of these matrices are both positive, and this gives the result. As a basic example for the above construction, we have the following matrix: 2 − N 2 ... 2  1  2 2 − N... 2  KN = √   N  ......  2 2 ... 2 − N We should mention that this matrix is in fact absolute almost Hadamard, in the real sense, as explained in [70]. There are many other interesting examples, coming from various block design constructions, and we refer here to [20]. HADAMARD MATRICES 147

Observe now that our basic example, namely the above matrix KN , is at the same time circulant and symmetric. We have in fact the following result, also from [20]:

∗ Theorem 9.19. Consider a circulant matrix H ∈ MN (R ), written Hij = γj−i. If the following conditions are satisfied, H is almost Hadamard, in a real sense: (1) The vector q = F ∗γ satisfies q ∈ TN . P ∗ (2) With ε = sgn(γ), ρi = r εrγi+r and ν = F ρ, we have ν > 0. Proof. We use the Fourier transform theory from Theorem 6.13 above. As a first observation, the orthogonality of U is equivalent to the condition (1). Regarding now the condition SU t > 0, this is equivalent to StU > 0. But: t X X X (S H)ij = SkiHkj = εi−kγj−k = εrγj−i+r = ρj−i k k r √ Thus StU is circulant, with ρ/ N as first row. We therefore have StU = F LF ∗ with ∗ t L = diag(ν) and ν = F ρ, so S U > 0 iff ν > 0, and we are done. See [20]. As an example here, consider the following vector, having length N = 2n + 1: q = (−1)n(1, −1, 1,..., −1, 1, 1, −1,..., 1, −1) This vector satisfies the conditions of Theorem 9.19, and produces the following circulant N × N real almost Hadamard matrix, from [20]:  −1 π −1 2π −1 (N−1)π  1 − cos N cos N ...... cos N  cos−1 (N−1)π 1 − cos−1 π ...... − cos−1 (N−2)π   N N N   −1 (N−2)π −1 (N−1)π −1 (N−3)π  1 − cos N cos N 1 ...... cos N  LN =  . . . .  N  . . . .     . . . .   . . . .  −1 π −1 2π −1 3π − cos N cos N − cos N ...... 1 We refer to the paper [20] for further details on all this, and for some other basic facts regarding the almost Hadamard matrices, in the real case. Some further analytic facts are available from [14], [15], [18], [70]. There is as well a concrete application of all this, to the minors of the Hadamard matrices, in the spirit of [65], available from [19].

Following now [17], let us discuss the complex case. Quite surprisingly, the above “basic” matrix√ KN is not an almost Hadamard matrix in the complex sense. That is, while KN / N locally maximizes the 1-norm on ON , it does not do so over UN . In fact, the same happens for the various matrices coming from Theorem 9.18 and Theorem 9.19 above. And, in addition to this, various complex matrices which can be constructed via straightforward complex extensions of the constructions in Theorem 9.18 and Theorem 9.19 fail to be almost Hadamard as well, in the complex sense. We are therefore led to the following statement, from [17]: 148 TEO BANICA

Conjecture 9.20 (Almost Hadamard conjecture, (AHC)). Any local maximizer of the 1-norm on UN must be a global maximizer, i.e. must be a rescaled Hadamard matrix. In other words, our conjecture would be that, in the complex setting, almost Hadamard implies Hadamard. This would be of course something very useful. Regarding now the known veriﬁcations of the AHC, as already mentioned above, these basically concern the natural “candidates” coming from Theorem 9.18 and Theorem 9.19, as well as some straightforward complex generalizations of these candidates. All this is quite technical, and generally speaking, we refer here to [17]. Let us mention, however, that the main idea that emerges from [17] is that of using a method based on a random derivative, pointing towards a suitable homogeneous space coset. In order to explain this, let OSCN ⊂ USCN be the orthogonal symmetric circulant matrices, and unitary self-adjoint circulant matrices. Via the Fourier transform inden- tiﬁcations from Theorem 6.13, the inclusion OSCN ⊂ USCN corresponds then to the (N+e)/2 N following inclusion, with Z2 with e = 0, 1 being {p ∈ Z2 |pk = p−k}: (N+e)/2 N Z2 ⊂ Z2

Let us consider as well the set USBN consisting of unitary bistochastic self-adjoint matrices. The various results in [17] suggest the following statement: ∗ Conjecture 9.21. Given U ∈ USBN satisfying S U ≥ 0, there exists a simple function B → BU , probably either the identity or the passage to another coset, such that Z Φ(U, BU )dB ≤ 0 OSC N √ and such that the equality can only be attained when H = NU is Hadamard. Observe that, in view of Theorem 9.17 above, this would more or less prove the AHC, modulo a remaining extension from USBN to the group UN itself. As already mentioned, this latter conjecture is supported by the computations in [17], which either use this idea, or can be reformulated in this spirit. Regarding the applications, the situation is of course very diﬀerent from the one in the real case. Assuming that the AHC holds indeed, we would have here a new approach to the complex Hadamard matrices, which is by construction analytic and local. This would be of course something quite powerful, potentially reshaping the whole subject. HADAMARD MATRICES 149

10. Quantum groups

We discuss in what follows the relation between the Hadamard matrices H ∈ MN (C) + and the quantum permutation groups G ⊂ SN , and its potential applications to certain mathematical physics questions. There is a lot of material to be surveyed here, and we will insist on mathematical aspects, regarding the correspondence H → G. We will need many preliminaries, first concerning the operator algebras, then the compact quantum groups in the sense of Woronowicz [99], [100], and then the matrix modelling questions for such quantum groups. Once done with this, we will be able to talk about quantum permutations, and their relation with the Hadamard matrices. Let H be a Hilbert space. We denote by B(H) the algebra of bounded operators T : H → H, with usual norm and involution. The algebra B(H), as well as any of its unital subalgebras A ⊂ B(H) which are complete, and stable under ∗, fit into: Definition 10.1. A unital C∗-algebra is a complex algebra with unit A, having: (1) A norm a → ||a||, making it a Banach algebra (the Cauchy sequences converge). (2) An involution a → a∗, which satisfies ||aa∗|| = ||a||2, for any a ∈ A. In what follows we will often omit the adjective “unital”, because the passage to the non-unital case would bring nothing interesting, in connection with our questions. Generally speaking, the elements a ∈ A are best thought of as being some kind of “generalized operators”, on some Hilbert space which is not present. By using this idea, one can emulate spectral theory in this setting, in the following way: Proposition 10.2. Given a ∈ A, define its spectrum as σ(a) = {λ ∈ C|a − λ 6∈ A−1}, and its spectral radius ρ(a) as the radius of the smallest centered disk containing σ(a). (1) The spectrum of a norm one element is in the unit disk. (2) The spectrum of a unitary element (a∗ = a−1) is on the unit circle. (3) The spectrum of a self-adjoint element (a = a∗) consists of real numbers. (4) The spectral radius of a normal element (aa∗ = a∗a) is equal to its norm. Proof. All this is standard, by using σ(f(a)) = f(σ(a)) for any f ∈ C[X], and more generally for any f ∈ C(X) having poles outside σ(a), which is elementary: 1 2 (1) This simply comes from 1−a = 1 + a + a + ... for any ||a|| < 1. (2) This follows from σ(a)−1 = σ(a−1) = σ(a∗) = σ(a). (3) This follows from (2), by using f(z) = (z + it)/(z − it), with t ∈ R. (4) We have ρ(a) ≤ ||a|| from (1). Conversely, given ρ > ρ(a), we have: ∞ Z zn X Z dz = zn−k−1dz ak = an−1 z − a |z|=ρ k=0 |z|=ρ n 1/n By applying the norm and taking n-th roots we obtain ρ ≥ limn→∞ ||a || , and then ∗ 2 by using ||aa || = ||a|| we obtain ρ ≥ ||a||, and so ρ(a) ≥ ||a||, as desired. 150 TEO BANICA

With these preliminaries in hand, we can now formulate some theorems. The basic facts about the C∗-algebras, that we will need here, can be summarized as: Theorem 10.3. The C∗-algebras have the following properties: (1) The commutative ones are those of the form C(X), with X compact space. (2) Any such algebra A embeds as A ⊂ B(H), for some Hilbert space H. (3) In finite dimensions, these are the direct sums of matrix algebras. Proof. All this is standard, the idea being as follows: (1) Given a compact space X, the algebra C(X) of continuous functions f : X → C ∗ ∗ is indeed a C -algebra, with norm ||f|| = supx∈X |f(x)|, and involution f (x) = f(x). Observe that this algebra is indeed commutative, because f(x)g(x) = g(x)f(x). Conversely, if A is commutative, we can define X = Spec(A) to be the space of characters χ : A → C, with topology making continuous all evaluation maps eva : χ → χ(a). We have then a morphism of algebras ev : A → C(X) given by a → eva, and Proposition 10.2 (3) shows that ev is a ∗-morphism, Proposition 10.2 (4) shows that ev is isometric, and finally the Stone-Weierstrass theorem shows that ev is surjective. (2) This is standard for A = C(X), where we can pick a probability measure on X, and set H = L2(X), and use the embedding A ⊂ B(H) given by f → (g → fg). In the general case, where A is no longer commutative, the proof is quite similar, by emulating basic measure theory in the abstract C∗-algebra setting. (3) Assuming that A is finite dimensional, we can first decompose its unit as 1 = p1 + ... + pk, with pi ∈ A being minimal projections. Each of the linear spaces Ai = piApi is then a non-unital ∗-subalgebra of A, and we have a non-unital ∗-algebra sum decomposition A = A1 ⊕ ... ⊕ Ak. On the other hand, since each pi is minimal, we have unital ∗-algebra isomorphisms Ai ' Mri (C), where ri = rank(pi). Thus, we obtain a ∗ C -algebra isomorphism A ' Mr1 (C) ⊕ ... ⊕ Mrk (C), as desired. All the above is of course quite brief, but details on all this can be found in any book on operator algebras. For a slightly longer proof of (1), called Gelfand theorem, and which is the key result that we will need here, we refer for instance to [12].

As a conclusion to all this, given a C∗-algebra A, we can think of it as being of the form A = C(X), with X being a “compact quantum space”. We will be interested here in the case where X is a “compact quantum group”. The axioms for the corresponding C∗-algebras, found by Woronowicz in [99], are, in a soft form, as follows: Definition 10.4. A Woronowicz algebra is a C∗-algebra A, given with a unitary matrix u ∈ MN (A) whose coefficients generate A, such that the formulae X ∗ ∆(uij) = uik ⊗ ukj , ε(uij) = δij ,S(uij) = uji k define morphisms of C∗-algebras ∆ : A → A ⊗ A, ε : A → C, S : A → Aopp. HADAMARD MATRICES 151

The morphisms ∆, ε, S are called comultiplication, counit and antipode. We say that A is cocommutative when Σ∆ = ∆, where Σ(a ⊗ b) = b ⊗ a is the ﬂip. We have the following result, which justiﬁes the terminology and axioms: Proposition 10.5. The following are Woronowicz algebras:

(1) C(G), with G ⊂ UN compact Lie group. Here the structural maps are: ∆(ϕ) = (g, h) → ϕ(gh) ε(ϕ) = ϕ(1) S(ϕ) = g → ϕ(g−1) ∗ (2) C (Γ), with FN → Γ ﬁnitely generated group. Here the structural maps are: ∆(g) = g ⊗ g ε(g) = 1 S(g) = g−1 Moreover, we obtain in this way all the commutative/cocommutative algebras. Proof. In both cases, we have to exhibit a certain matrix u. For the ﬁrst assertion, we can use the matrix u = (uij) formed by matrix coordinates of G, given by:

 u11(g) . . . u1N (g)  . . g =  . .  uN1(g) . . . uNN (g) For the second assertion, we can use the diagonal matrix formed by generators:

g1 0  . u =  ..  0 gN Finally, the last assertion follows from the Gelfand theorem, in the commutative case, and in the cocommutative case, this follows from the results of Woronowicz in [99]. In general now, the structural maps ∆, ε, S have the following properties: Proposition 10.6. Let (A, u) be a Woronowicz algebra. (1) ∆, ε satisfy the usual axioms for a comultiplication and a counit, namely: (∆ ⊗ id)∆ = (id ⊗ ∆)∆ (ε ⊗ id)∆ = (id ⊗ ε)∆ = id (2) S satisﬁes the antipode axiom, on the ∗-subalgebra generated by entries of u: m(S ⊗ id)∆ = m(id ⊗ S)∆ = ε(.)1 (3) In addition, the square of the antipode is the identity, S2 = id. 152 TEO BANICA

Proof. The two comultiplication axioms follow from: X (∆ ⊗ id)∆(uij) = (id ⊗ ∆)∆(uij) = uik ⊗ ukl ⊗ ulj kl

(ε ⊗ id)∆(uij) = (id ⊗ ε)∆(uij) = uij As for the antipode formulae, the verification here is similar. Summarizing, the Woronowicz algebras appear to have very nice properties. In view of Proposition 10.5 above, we can now formulate the following definition: Definition 10.7. Given a Woronowicz algebra A, we formally write A = C(G) = C∗(Γ) and call G compact quantum group, and Γ discrete quantum group. When A is both commutative and cocommutative, G is a compact abelian group, Γ is a discrete abelian group, and these groups are dual to each other, G = Γb, Γ = Gb. In general, we still agree to write G = Γb, Γ = Gb, but in a formal sense. With this in mind, let us call now corepresentation of A any unitary matrix v ∈ Mn(A) satisfying the same conditions are those satisfied by u, namely:

X ∗ ∆(vij) = vik ⊗ vkj , ε(vij) = δij ,S(vij) = vji k These corepresentations can be thought of as corresponding to the unitary representations of the underlying compact quantum group G. As main examples, we have u = (uij) ∗ itself, its conjugateu ¯ = (uij), as well as any tensor product between u, u¯. We have the following key result, due to Woronowicz [99]: Theorem 10.8. Any Woronowicz algebra has a unique Haar integration functional, Z Z Z ⊗id ∆ = id ⊗ ∆ = (.)1 G G G which can be constructed by starting with any faithful positive form ϕ ∈ A∗, and setting n Z 1 X = lim ϕ∗k n→∞ n G k=1 where φ ∗ ψ = (φ ⊗ ψ)∆. Moreover, for any corepresentation v ∈ Mn(C) ⊗ A we have Z id ⊗ v = P G where P is the orthogonal projection onto F ix(v) = {ξ ∈ Cn|vξ = ξ}. HADAMARD MATRICES 153

Proof. Following [99], this can be done in 3 steps, as follows: (1) Given ϕ ∈ A∗, our claim is that the following limit converges, for any a ∈ A: n Z 1 X a = lim ϕ∗k(a) n→∞ n ϕ k=1 Indeed, by linearity we can assume that a is the coeﬃcient of corepresentation, a = (τ ⊗ id)v. But in this case, an elementary computation shows that we have the following formula, where Pϕ is the orthogonal projection onto the 1-eigenspace of (id ⊗ ϕ)v: Z id ⊗ v = Pϕ ϕ

(2) Since vξ = ξ implies [(id ⊗ ϕ)v]ξ = ξ, we have Pϕ ≥ P , where P is the orthogonal projection onto the space F ix(v) = {ξ ∈ Cn|vξ = ξ}. The point now is that when ϕ ∈ A∗ is faithful, by using a positivity trick, one can prove that we have Pϕ = P . Thus our R linear form ϕ is independent of ϕ, and is given on coeﬃcients a = (τ ⊗ id)v by: Z id ⊗ v = P ϕ R R (3) With the above formula in hand, the left and right invariance of G = ϕ is clear on coeﬃcients, and so in general, and this gives all the assertions. See [99]. The above result is something quite fundamental, and as a main application, one can develop a Peter-Weyl type theory for the corepresentations of A. See [99].

Finally, we will need some general theory regarding the random matrix models for the Woronowicz algebras. The idea here is very simple, namely that of modelling the coordinates uij ∈ A by certain concrete variables Uij ∈ B. Our favorite type of algebras being the random matrix ones, B = MK (C(T )), we are led into: Deﬁnition 10.9. A matrix model for A = C(G) is a morphism of C∗-algebras

π : C(G) → MK (C(T )) where T is a compact space, and K ≥ 1 is an integer.

The “best” models are of course the faithful ones, π : C(G) ⊂ MK (C(T )). However, this formalism is quite restrictive, not covering many interesting examples. In order to fix this, let us look at the group dual case, A = C∗(Γ), with Γ being a usual ∗ discrete group. We know that a model π : C (Γ) → MK (C(T )) must come from a group representation ρ :Γ → C(T,UK ). Now observe that when ρ is faithful, the representation π is in general not faithful, for instance because when T = {.} its target algebra is finite dimensional. On the other hand, this representation “reminds” Γ, and so can be used in order to fully understand Γ. This leads to the following definition: 154 TEO BANICA

Deﬁnition 10.10. Let π : C(G) → MK (C(T )) be a matrix model. (1) The Hopf image of π is the smallest quotient Hopf C∗-algebra C(G) → C(H) producing a factorization of type π : C(G) → C(H) → MK (C(T )). (2) When the inclusion H ⊂ G is an isomorphism, i.e. when there is no non-trivial factorization as above, we say that π is inner faithful.

In the case where G = Γb is a group dual, π must come from a group representation ρ :Γ → C(T,UK ), and the above factorization is simply the one obtained by taking the image, ρ :Γ → Λ ⊂ C(T,UK ). Thus π is inner faithful when Γ ⊂ C(T,UK ). Also, given a compact group G, and elements g1, . . . , gK ∈ G, we have a representation K π : C(G) → C , given by f → (f(g1), . . . , f(gK )). The minimal factorization of π is then via C(H), with H = < g1, . . . , gK >, and π is inner faithful when G = H. In general, the existence and uniqueness of the Hopf image comes from dividing C(G) by a suitable ideal. We refer to [11] for more details regarding this construction.

We will be interested here in the quantum permutation groups, and their relation with the Hadamard matrices. The following key deﬁnition is due to Wang [96]: Proposition 10.11. A magic unitary matrix is a square matrix over a C∗-algebra,

u ∈ MN (A) whose entries are projections, summing up to 1 on each row and each column.

The basic examples of such matrices come from the usual permutation groups, G ⊂ SN . Indeed, given such subgroup, the following matrix is magic:

uij = χ σ ∈ G σ(j) = i This leads us into the following key deﬁnition, due to Wang [96] as well:

+ ∗ Definition 10.12. C(SN ) is the universal C -algebra generated by the entries of a N ×N magic unitary matrix u = (uij), with the morphisms given by X ∆(uij) = uik ⊗ ukj , ε(uij) = δij ,S(uij) = uji k as comultiplication, counit and antipode. This algebra satisfies the axioms in Definition 10.4, and the underlying compact quan- + tum group SN is called quantum permutation group. Quite surprisingly, we have: + Theorem 10.13. We have an embedding SN ⊂ SN , given at the algebra level by:

uij → χ σ σ(j) = i + This is an isomorphism at N ≤ 3, but not at N ≥ 4, where SN is not classical, nor ﬁnite. HADAMARD MATRICES 155

Proof. The fact that we have indeed an embedding as above is clear. Regarding now the second assertion, we can prove this in four steps, as follows: + Case N = 2. The fact that S2 is indeed classical, and hence collapses to S2, is trivial, because the 2 × 2 magic matrices are as follows, with p being a projection: p 1 − p U = 1 − p p

Case N = 3. It is enough to check that u11, u22 commute. But this follows from:

u11u22 = u11u22(u11 + u12 + u13)

= u11u22u11 + u11u22u13

= u11u22u11 + u11(1 − u21 − u23)u13

= u11u22u11 Indeed, by applying the involution to this formula, we obtain from this that we have u22u11 = u11u22u11 as well, and so we get u11u22 = u22u11, as desired. Case N = 4. Consider the following matrix, with p, q being projections:  p 1 − p 0 0  1 − p p 0 0  U =    0 0 q 1 − q 0 0 1 − q q This matrix is then magic, and if we choose p, q as for the algebra < p, q > to be infinite + dimensional, we conclude that C(S4 ) is infinite dimensional as well. + + Case N ≥ 5. Here we can use the standard embedding S4 ⊂ SN , obtained at the level of the corresponding magic matrices in the following way: u 0 u → 0 1N−4 + Indeed, with this in hand, the fact that S4 is a non-classical, infinite compact quantum + group implies that SN with N ≥ 5 has these two properties as well. See [96]. + −1 At a more advanced level, one can prove that S4 ' SO3 . At N ≥ 5 the quantum + group SN still has the same fusion rules as SO3, but is not coamenable. See [12]. In relation now with the complex Hadamard matrices, the connection with the quantum permutations is immediate, coming from the following observation:

Proposition 10.14. If H ∈ MN (C) is Hadamard, the rank one projections Hi Pij = P roj Hj N where H1,...,HN ∈ T are the rows of H, form a magic unitary. 156 TEO BANICA

Proof. This is clear, the verification for the rows being as follows: Hi Hi X Hil Hkl X Hkl , = · = = Nδjk Hj Hk Hjl Hil Hjl l l The verification for the columns is similar, we follows: Hi Hk X Hil Hjl X Hil , = · = = Nδik Hj Hj Hjl Hkl Hkl l l Thus, we have indeed a magic unitary, as claimed. Summarizing, any complex Hadamard matrix produces a representation of the quan- + tum permutation algebra C(SN ). Thus, we can apply the Hopf image construction from Definition 10.10, and we are led in this way into the following notion:

Deﬁnition 10.15. To any Hadamard matrix H ∈ MN (C) we associate the quantum + permutation group G ⊂ SN given by the following Hopf image factorization,

+ π C(SN ) / MN (C) :

$ C(G)

N where π(uij) = P roj(Hi/Hj), with H1,...,HN ∈ T being the rows of H. Our claim now is that this construction H → G is something really useful, with G encoding the combinatorics of H. To be more precise, philosophically speaking, the idea will be that “H can be thought of as being a kind of Fourier matrix for G”. There are several results supporting this, with the main evidence coming from the following result, which collects the basic known results regarding the construction: Theorem 10.16. The construction H → G has the following properties:

(1) For a Fourier matrix H = FG we obtain the group G itself, acting on itself. (2) For H 6∈ {FG}, the quantum group G is not classical, nor a group dual. (3) For a tensor product H = H0 ⊗ H00 we obtain a product, G = G0 × G00. Proof. All this material is standard, and elementary, as follows: (1) Let us ﬁrst discuss the cyclic group case, H = FN . Here the rows of H are given i 2 N−1 by Hi = ρ , where ρ = (1, w, w , . . . , w ). Thus, we have the following formula: H i = ρi−j Hj

It follows that the corresponding rank 1 projections Pij = P roj(Hi/Hj) form a circulant matrix, all whose entries commute. Since the entries commute, the corresponding HADAMARD MATRICES 157 quantum group must satisfy G ⊂ SN . Now by taking into account the circulant property of P = (Pij) as well, we are led to the conclusion that we have G = ZN . In the general case now, where H = FG, with G being an arbitrary ﬁnite abelian group, the result can be proved either by extending the above proof, of by decomposing

G = ZN1 × ... × ZNk and using (3) below, whose proof is independent from the rest. (2) This is something more tricky, needing some general study of the representations whose Hopf images are commutative, or cocommutative. For details here, along with a number of supplementary facts on the construction H → G, we refer to [13], [21]. (3) Assume that we have a tensor product H = H0 ⊗ H00, and let G, G0,G00 be the associated quantum permutation groups. We have then a diagram as follows:

+ + 0 00 C(SN 0 ) ⊗ C(SN 00 ) / C(G ) ⊗ C(G ) / MN 0 (C) ⊗ MN 00 (C) O

+ C(SN ) / C(G) / MN (C) Here all the maps are the canonical ones, with those on the left and on the right coming from N = N 0N 00. At the level of standard generators, the diagram is as follows:

0 00 0 00 0 00 uij ⊗ uab / wij ⊗ wab / Pij ⊗ Pab O

uia,jb / wia,jb / Pia,jb Now observe that this diagram commutes. We conclude that the representation asso- 0 00 ciated to H factorizes indeed through C(G ) ⊗ C(G ), and this gives the result. Generally speaking, going beyond Theorem 10.16 is a quite diﬃcult question. There are several computations available here, for the most regarding the deformations of the Fourier matrices, and we will be back to all this later on, in section 12 below.

We would like to end this section with two theoretical extensions of the construction H → G from Deﬁnition 10.15, which are both quite interesting. A ﬁrst idea, from [10], is that of using complex Hadamard matrices with noncommutative entries. Consider an arbitrary unital C∗-algebra A. Two row or column vectors over this algebra, say a = (a1, . . . , aN ) and b = (b1, . . . , bN ), are called orthogonal when:

X ∗ X ∗ aibi = ai bi = 0 i i P ∗ P ∗ Observe that by applying the involution, we have as well i biai = i bi ai = 0. 158 TEO BANICA

With this notion in hand, we can formulate:

Deﬁnition 10.17. An Hadamard matrix over a unital C∗-algebra A is a square matrix H ∈ MN (A) satisfying the following conditions:

(1) All the entries of H are unitaries, Hij ∈ UA. (2) These entries commute on all rows and all columns of H. (3) The rows and columns of H are pairwise orthogonal.

As a ﬁrst remark, in the simplest case A = C the unitary group is the unit circle in the complex plane, UC = T, and we obtain the usual complex Hadamard matrices. In the general commutative case, A = C(X), our Hadamard matrix must be made of “ﬁbers”, one for each point x ∈ X. Thus, we must have H = {Hx|x ∈ X}, with Hx being complex Hadamard matrices, depending continuously on x ∈ X. When A is not commutative, we can have many interesting examples, which can be quite far away from the usual Hadamard matrices. We will be back to this later. In general now, observe that if H = (Hij) is Hadamard, then so are the matrices ¯ ∗ t ∗ ∗ H = (Hij), H = (Hji) and H = (Hji). In addition, we have the following result:

Proposition 10.18. The class of Hadamard matrices H ∈ MN (A) is stable under: (1) Permuting the rows or columns. (2) Multiplying the rows or columns by central unitaries. When successively combining these two operations, we obtain an equivalence relation.

Proof. This is clear indeed from deﬁnitions.

Observe that in the commutative case A = C(X) any unitary is central, so we can multiply the rows or columns by any unitary. In particular in this case we can always “dephase” the matrix, i.e. assume that its ﬁrst row and column consist of 1 entries. Let us discuss now the tensor products, and their deformations. Following [47], the deformed tensor products are constructed as follows:

Theorem 10.19. Let H ∈ MN (A) and K ∈ MM (A) be Hadamard matrices, and Q ∈ MN×M (UA). Then the “deformed tensor product” H ⊗Q K ∈ MNM (A), given by

(H ⊗Q K)ia,jb = QibHijKab is an Hadamard matrix as well, provided that the entries of Q commute on rows and columns, and that the algebras < Hij >, < Kab >, < Qib > pairwise commute. HADAMARD MATRICES 159

Proof. First, the entries of L = H ⊗Q K are unitaries, and its rows are orthogonal:

X ∗ X ∗ ∗ ∗ Lia,jbLkc,jb = QibHijKab · QkbKcbHkj jb jb X ∗ ∗ = Nδik QibKab · QkbKcb b X ∗ = Nδik KabKcb j

= NM · δikδac The orthogonality of columns can be checked as follows:

X ∗ X ∗ ∗ ∗ Lia,jbLia,kc = QibHijKab · QicKacHik ia ia X ∗ ∗ = Mδbc QibHij · QicHik i X ∗ = Mδbc HijHik i = NM · δjkδbc For the commutation on rows we use in addition the commutation on rows for Q:

Lia,jbLkc,jb = QibHijKab · QkbHkjKcb

= QibQkb · HijHkj · KabKcb

= QkbQib · HkjHij · KcbKab

= QkbHkjKcb · QibHijKab

= Lkc,jbLia,jb The commutation on columns is similar, using the commutation on columns for Q:

Lia,jbLia,kc = QibHijKab · qicHikKac

= QibQic · HijHik · KabKac

= QicQib · HikHij · KacKab

= QicHikKac · QibHijKab

= Lia,kcLia,jb

Thus all the axioms are satisﬁed, and L is indeed Hadamard.

As a basic example, we have the following construction: 160 TEO BANICA

Proposition 10.20. The following matrix is Hadamard,

x y x y  x −y x −y M =   z t −z −t z −t −z t for any unitaries x, y, z, t satisfying [x, y] = [x, z] = [y, t] = [z, t] = 0.

Proof. This follows indeed from Theorem 10.19, because we have:

x y x y  1 1 1 1 x −y x −y ⊗  =   1 −1 x y 1 −1 z t −z −t z t z −t −z t

In addition, the commutation relations in Theorem 10.19 are satisﬁed indeed.

The generalized Hadamard matrices produce quantum groups, as follows:

Theorem 10.21. If H ∈ MN (A) is Hadamard, the following matrices Pij ∈ MN (A) form altogether a magic matrix P = (Pij), over the algebra MN (A):

1 (P ) = H H∗ H H∗ ij ab N ia ja jb ib

+ Thus, we can let π : C(SN ) → MN (A) be the representation associated to P , and then + + factorize π : C(SN ) → C(G) → MN (A), with G ⊂ SN chosen minimal.

Proof. The magic condition can be checked in three steps, as follows: ∗ 2 (1) Let us ﬁrst check that each Pij is a projection, i.e. that we have Pij = Pij = Pij. ∗ Regarding the ﬁrst condition, namely Pij = Pij, this simply follows from:

1 (P )∗ = (H H∗ H H∗ )∗ ij ba N ib jb ja ia 1 = H H∗ H H∗ N ia ja jb ib = (Pij)ab HADAMARD MATRICES 161

2 As for the second condition, Pij = Pij, this follows from the fact that all the entries Hij are assumed to be unitaries, i.e. follows from axiom (1) in Deﬁnition 10.17:

2 X (Pij)ab = (Pij)ac(Pij)cb c 1 X = H H∗ H H∗ H H∗ H H∗ N 2 ia ja jc ic ic jc jb ib c 1 = H H∗ H H∗ N ia ja jb ib = (Pij)ab (2) Let us check now that fact that the entries of P sum up to 1 on each row. For this ∗ purpose we use the equality H H = N1N , coming from the axiom (3), which gives: X 1 X ( P ) = H H∗ H H∗ ij ab N ia ja jb ib j j 1 = H (H∗H) H∗ N ia ab ib ∗ = δabHiaHib

= δab (3) Finally, let us check that the entries of P sum up to 1 on each column. This is t ¯ the trickiest check, because it involves, besides axiom (1) and the formula H H = N1N coming from axiom (3), the commutation on the columns of H, coming from axiom (2): X 1 X ( P ) = H H∗ H H∗ ij ab N ia ja jb ib i i 1 X = H∗ H H∗ H N ja ia ib jb i 1 = H∗ (HtH¯ ) H N ja ab jb ∗ = δabHjaHjb

= δab

Thus P is indeed a magic matrix in the above sense, and we are done.

As an illustration, consider a usual Hadamard matrix H ∈ MN (C). If we denote its rows by H1,...,HN and we consider the vectors ξij = Hi/Hj, then we have: Hi1 HiN ξij = ,..., Hj1 HjN 162 TEO BANICA

Thus the orthogonal projection on this vector ξij is given by:

1 1 ∗ ∗ (Pξij )ab = 2 (ξij)a(ξij)b = HiaHjaHjbHib = (Pij)ab ||ξij|| N

We conclude that we have Pij = Pξij for any i, j, so our construction from Theorem 10.21 is compatible with the construction for the usual complex Hadamard matrices.

In general, computing G is a quite diﬃcult question, and the answer for instance for the matrices in Proposition 10.20 is not known. We refer to [10] for more on this.

Let us discuss now another generalization of the construction H → G. The idea, following [23], will be that of looking at the partial Hadamard matrices (PHM), and their connection with the partial permutations. Let us start with: Deﬁnition 10.22. A partial permutation of {1 ...,N} is a bijection σ : X ' Y , with X,Y ⊂ {1,...,N}. We denote by SeN the set formed by such partial permutations.

Observe that we have SN ⊂ SeN . The embedding u : SN ⊂ MN (0, 1) given by permutation matrices can be extended to an embedding u : SeN ⊂ MN (0, 1), as follows: ( 1 if σ(j) = i u (σ) = ij 0 otherwise

By looking at the image of this embedding, we see that SeN is in bijection with the matrices M ∈ MN (0, 1) having at most one 1 entry on each row and column. In analogy with Wang’s theory in [96], we have the following deﬁnition:

Deﬁnition 10.23. A submagic matrix is a matrix u ∈ MN (A) whose entries are projec- + tions, which are pairwise orthogonal on rows and columns. We let C(SeN ) be the universal C∗-algebra generated by the entries of a N × N submagic matrix.

+ P The algebra C(SeN ) has a comultiplication given by ∆(uij) = k uik ⊗ukj, and a counit + given by ε(uij) = δij. Thus SeN is a quantum semigroup, and we have maps as follows, with the bialgebras at left corresponding to the quantum semigroups at right:

+ + + + C(SeN ) → C(SN ) SeN ⊃ SN

↓ ↓ : ∪ ∪

C(SeN ) → C(SN ) SeN ⊃ SN The relation of all this with the PHM is immediate, appearing as follows: HADAMARD MATRICES 163

N Theorem 10.24. If H ∈ MM×N (T) is a PHM, with rows denoted H1,...,HM ∈ T , then the following matrix of rank one projections is submagic: Hi Pij = P roj Hj + Thus H produces a representation πH : C(SeM ) → MN (C), given by uij → Pij, that we + can factorize through C(G), with the quantum semigroup G ⊂ SeM chosen minimal. Proof. We have indeed the following computation, for the rows:

D Hi Hi E X Hil Hkl X Hkl , = · = =< Hk,Hj >= δjk Hj Hk Hjl Hil Hjl l l The veriﬁcation for the columns is similar, we follows: Hi Hk X Hil Hjl X Hil , = · = = Nδik Hj Hj Hjl Hkl Hkl l l Regarding now the last assertion, we can indeed factorize our representation as indicated, with the existence and uniqueness of the bialgebra C(G), with the minimality + property as above, being obtained by dividing C(SeM ) by a suitable ideal. See [23]. The very ﬁrst problem is that of deciding under which exact assumptions our construction is in fact “classical”. In order to explain the answer here, we will need:

Deﬁnition 10.25. A pre-Latin square is a matrix L ∈ MM (1,...,N) having the property that its entries are distinct, on each row and each column. Given such a matrix L, to any x ∈ {1,...,N} we can associate the partial permutation σx ∈ SeM given by σx(j) = i ⇐⇒ Lij = x. We denote by G ⊂ SeM the semigroup generated by σ1, . . . , σN , and call it semigroup associated to L. N Also, given an orthogonal basis ξ = (ξ1, . . . , ξN ) of C , we can construct a submagic matrix P ∈ MM (MN (C)), according to the formula Pij = P roj(ξLij ). With these notations, we have the following result, from [23]:

Theorem 10.26. If H ∈ MN×M (C) is a PHM, the following are equivalent: + (1) The semigroup G ⊂ SeM is classical, i.e. G ⊂ SeM . (2) The projections Pij = P roj(Hi/Hj) pairwise commute. N (3) The vectors Hi/Hj ∈ T are pairwise proportional, or orthogonal. (4) The submagic matrix P = (Pij) comes for a pre-Latin square L. In addition, if so is the case, G is the semigroup associated to L. Proof. Here (1) ⇐⇒ (2) is clear, (2) ⇐⇒ (3) comes from the fact that two rank 1 projections commute precisely when their images coincide, or are orthogonal, (3) ⇐⇒ (4) is clear again, and the last assertion comes from Gelfand duality. See [23]. 164 TEO BANICA

We call “classical” the matrices in Theorem 10.26. There are many examples here, the most basic ones being the upper M × N submatrices of the Fourier matrices FN , denoted FM,N . If we denote by GM,N ⊂ SeM the associated semigroups, we have:

Theorem 10.27. In the N > 2M − 2 regime, GM,N ⊂ SeM is formed by the maps •••• • • • σ = •••••••Ð Ð Ð that is, σ : I ' J, σ(j) = j − x, with I,J ⊂ {1,...,M} intervals, independently of N.

Proof. Since for He = FN the associated Latin square is circulant, Leij = j−i, the pre-Latin square that we are interested in is:  0 1 2 ...M − 1  N − 1 0 1 ...M − 2   L =  N − 2 N − 1 0 ...M − 3    ...  N − M + 1 N − M + 2 N − M + 3 ... 0 Observe that, due to our N > 2M − 2 assumption, we have N − M + 1 > M − 1, and so the entries above the diagonal are distinct from those below the diagonal. With this remark in hand, the computation is quite standard. See [22].

In the remaining regime, M < N ≤ 2M − 2, the semigroup GM,N ⊂ SeM looks quite hard to compute, and for the moment we only have some partial results regarding it. For a partial permutation σ : I ' J with |I| = |J| = k, set κ(σ) = k. We have:

(k) Theorem 10.28. The components GM,N = {σ ∈ GM,N |κ(σ) = k} with k > 2M − N are, in the M < N ≤ 2M − 2 regime, the same as those in the N > 2M − 2 regime. Proof. The pre-Latin square that we are interested in has as usual 0 on the diagonal, and then takes its entries from the set S = {1,...,N − M} ∪ {N − M + 1,...,M − 1} ∪ {M,...,N − 1}, in a uniform way from each of the 3 components of S. With this remark in hand, the proof is quite standard. See [22]. There are many interesting questions regarding the construction H → G, in this generalized PHM/partial permutation setting, and we refer here to [22], [23]. HADAMARD MATRICES 165

11. Subfactor theory We discuss here some potential applications of the construction H → G, and of the Hadamard matrices in general, to certain questions from mathematical physics. Generally speaking, all this is related to statistical mechanics. The idea indeed, which is old folklore, is that associated to any 2D spin model should be a quantum permutation + + group G ⊂ SN , which appears by factorizing the ﬂat representation C(SN ) → MN (C) associated to the N × N matrix of the Boltzmann weights of the model, and whose representation theory computes the partition function of the model. All this comes from the work of Jones in subfactor theory [59], [60], [61], and from various general correspondences between quantum groups and subfactors. There are some direct computations as well, supporting this idea, as those in [16]. However, all this is not axiomatized yet. So, as a more modest goal here, we will explain the relation between the Hadamard matrices and the von Neumann algebras, the commuting squares, the subfactor theory, and Jones’ planar algebra work in [61], and then we will comment on some further possible developments of all this.

In order to start, we will need some basic von Neumann algebra theory, coming as a complement to the basic C∗-algebra theory explained in section 10 above: Theorem 11.1. The von Neumann algebras, which are the C∗-algebras A ⊂ B(H) closed under the weak topology, making each T → T x continuous, are as follows: (1) They are exactly the ∗-algebras of operators A ⊂ B(H) which are equal to their bicommutant, A = A00. (2) In the commutative case, these are the algebras of type A = L∞(X), with X measured space, represented on H = L2(X), up to a multiplicity. (3) If we write the center as Z(A) = L∞(X), then we have a decomposition of type R A = X Ax dx, with the fibers Ax having trivial center, Z(Ax) = C. (4) The factors, Z(A) = C, can be fully classified in terms of II1 factors, which are those satisfying dim A = ∞, and having a faithful trace tr : A → C. (5) The II1 factors enjoy the “continuous dimension geometry” property, in the sense that the traces of their projections can take any values in [0, 1]. (6) Among the II1 factors, the most important one is the Murray-von Neumann hy- perfinite factor R, obtained as an inductive limit of matrix algebras. Proof. This is something quite heavy, the idea being as follows: (1) This is von Neumann’s bicommutant theorem, which is well-known in finite dimensions, and whose proof in general is not that complicated, either. (2) It is clear, via basic measure theory, that L∞(X) is indeed a von Neumann algebra on H = L2(X). The converse can be proved as well, by using spectral theory. (3) This is von Neumann’s reduction theory main result, whose statement is already quite hard to understand, and whose proof uses advanced functional analysis. 166 TEO BANICA

(4) This is something heavy, due to Murray-von Neumann and Connes, the idea being that the other factors can be basically obtained via crossed product constructions. (5) This is a jewel of functional analysis, with the rational traces being relatively easy to obtain, and with the irrational ones coming from limiting arguments. (6) Once again, heavy results, due to Murray-von Neumann and Connes, the idea being that any ﬁnite dimensional construction always leads to the same factor, called R. All the above is of course very brief. We recommend here the original papers of von Neumann and Connes, starting for instance with [71], and then [37].

As a philosophical comment, observe the huge technical diﬀerence between the basic C∗- algebra theory, more or less explained in section 10 above, and the basic von Neumann algebra theory, barely discussed above. Some theories are much more advanced than others, perhaps because they are more interesting, or more beautiful, or both. As a side remark here, in view of all this, it would have been of course desirable to introduce the compact quantum groups G by talking directly about the associated von Neumann algebras L∞(G). Unfortunately this is not possible, because the underlying Hilbert spaces H = L2(G) do not come “by deﬁnition”, but by theorem. In addition, we cannot really talk about von Neumann algebras with generators and relations. In short, there is some philosophical clash here, between C∗-algebra theory and von Neumann algebra theory. This is a bit like a dispute between topology and probability. You don’t really need open and closed sets in order to do interesting mathematics, but if you have the occasion of learning some, why not having them in your bag of tricks.

From a more relaxed perspective, all this can be traced back to the Bohr-Einstein debates, the main question being whether God plays dice or not.

In relation now with our questions, variations of von Neumann’s reduction theory idea, basically using the abelian subalgebra Z(A) ⊂ A, include the use of maximal abelian subalgebras B ⊂ A, called MASA. In the finite von Neumann algebra case, where we have a trace, the use of orthogonal MASA is a standard method as well: Definition 11.2. A pair of orthogonal MASA is a pair of maximal abelian subalgebras B,C ⊂ A which are orthogonal with respect to the trace, (B C1) ⊥ (C C1). Here the scalar product is by definition < b, c >= tr(bc∗), and by taking into account the multiples of the identity, the orthogonality condition reformulates as follows: tr(bc) = tr(b)tr(c) , ∀b ∈ B, c ∈ C This notion is potentially useful in the infinite dimensional context, in relation with various structure and classification problems for the II1 factors. However, as a “toy HADAMARD MATRICES 167 example”, we can try and see what happens for the simplest factor that we know, namely the matrix algebra MN (C), with N ∈ N, endowed with its usual matrix trace. In this context, we have the following surprising observation of Popa [79]: Theorem 11.3. Up to a conjugation by a unitary, the pairs of orthogonal MASA in the simplest factor, namely the matrix algebra MN (C), are as follows, A = ∆ ,B = H∆H∗ with ∆ ⊂ MN (C) being the diagonal matrices, and with H ∈ MN (C) being Hadamard.

Proof. Any MASA in MN (C) being conjugated to ∆, we can assume, up to conjugation ∗ by a unitary, that we have A = ∆ and B = U∆U , with U ∈ UN . Now observe that given two diagonal matrices D,E ∈ ∆, we have: 1 X tr(D · UEU ∗) = (DUEU ∗) N ii i 1 X = D U E U¯ N ii ij jj ij ij 1 X = D E |U |2 N ii jj ij ij Thus, the orthogonality condition A ⊥ B reformulates as follows: 1 X 1 X D E |U |2 = D E N ii jj ij N 2 ii jj ij ij

But this tells us precisely that the entries |Uij√| must have the same absolute value, namely √1 , and so that the rescaled matrix H = NU must be Hadamard. N The above result is something quite fascinating, and in stark contrast with the mathematical solidity and beauty of Theorem 11.1. Bluntly put, there is a “black hole” in the foundations of modern von Neumann algebra theory, produced by their complex Hada- mard matrices, and their wild structure, geometry and combinatorics. Whether this black hole must be studied a bit, run away from, or simply ignored, is one of the main philosophical questions in modern von Neumann algebra theory. Generally speaking, subfactor theory and related areas are in favor of the “study a bit” direction, while free probability and related areas opt for the “run away from” solution. In relation with this, the present text is based on a bit of a hybrid philosophy, namely study the complex Hadamard matrices, of course and deﬁnitely, but with the idea in mind of eventually reaching to tools coming from Voiculescu’s free probability theory.

Getting back now to work, and to Theorem 11.3 above, as it is, along the same lines, but at a more advanced level, we have the following result: 168 TEO BANICA

Theorem 11.4. Given a complex Hadamard matrix H ∈ MN (C), the diagram formed by the associated pair of orthogonal MASA, namely

∆ / MN (C) O O

C / H∆H∗ is a commuting square in the sense of subfactor theory, in the sense that the expectations onto ∆,H∆H∗ commute, and their product is the expectation onto C.

Proof. It follows from deﬁnitions that the expectation E∆ : MN (C) → ∆ is the operation M → M∆ which consists in keeping the diagonal, and erasing the rest. ∗ Regarding now the other expectation, EH∆H∗ : MN (C) → H∆H ,√ it is better to ∗ identify it with the expectation EU∆U ∗ : MN (C) → U∆U , with U = H/ N. This latter ∗ expectation must be given by a formula of type M → UX∆U , with X satisfying:

∗ ∗ ∗ < M, UDU >=< UX∆U ,UDU > , ∀D ∈ ∆

The scalar products being given by < a, b >= tr(ab∗), this condition reads:

∗ ∗ ∗ tr(MUD U ) = tr(X∆D ) , ∀D ∈ ∆

Thus X = U ∗MU, and the formulae of our two expectations are as follows:

E∆(M) = M∆ ∗ ∗ EU∆U ∗ (M) = U(U MU)∆U

With these formulae in hand, we have the following computation:

∗ ∗ (E∆EU∆U ∗ M)ij = δij(U(U MU)∆U )ii X ∗ ¯ = δij Uik(U MU)kkUik k X 1 = δ · (U ∗MU) ij N kk k ∗ = δijtr(U MU)

= δijtr(M)

= (ECM)ij HADAMARD MATRICES 169

As for the other composition, the computation here is similar, as follows: ∗ ∗ (EU∆U ∗ E∆M)ij = (U(U M∆U)∆U )ij X ∗ ¯ = Uik(U M∆U)kkUjk k X ¯ ¯ = UikUlkMllUlkUjk kl 1 X = U M U¯ N ik ll jk kl

= δijtr(M)

= (ECM)ij Thus, we have indeed a commuting square, as claimed. We should mention that the notion of commuting square, which was heavily used by Popa in his classiﬁcation work for the subfactors [80], is a bit more complicated that what was said above. However, the other axioms are trivially satisﬁed for the class of commuting squares from Theorem 11.4, so we will not get into this. See [80].

As a conclusion, all this leads us into subfactor theory. So, let us explain now, following Jones [59], the basic theory here. Given an inclusion of II1 factors A0 ⊂ A1, which is actually something quite natural in physics, we can consider the orthogonal projection e1 : A1 → A0, and set A2 =< A1, e1 >. This procedure, called “basic construction”, can be iterated, and we obtain in this way a whole tower of II1 factors, as follows:

A0 ⊂e1 A1 ⊂e2 A2 ⊂e3 A3 ⊂ ...... The basic construction is something quite subtle, making deep connections with advanced mathematics and physics. All this was discovered by Jones in 1983, and his main result from [59], which came as a big surprise at that time, along with some supplementary fundamental work, done later, in [61], can be summarized as follows:

Theorem 11.5. Let A0 ⊂ A1 be an inclusion of II1 factors.

(1) The sequence of projections e1, e2, e3,... ∈ B(H) produces a representation of the Temperley-Lieb algebra TLN ⊂ B(H), where N = [A1,A0]. 0 (2) The collection P = (Pk) of the linear spaces Pk = A0 ∩ Ak, which contains the image of TLN , has a planar algebra structure. (3) The index N = [A1,A0], which is a Murray-von Neumann continuous quantity 2 π N ∈ [1, ∞], must satisfy N ∈ {4 cos ( n )|n ∈ N} ∪ [4, ∞]. Proof. This is something quite heavy, the idea being as follows: (1) The idea here is that the functional analytic study of the basic construction leads to the conclusion that the sequence of projections e1, e2, e3,... ∈ B(H) behaves algebrically 170 TEO BANICA

∪ ∪ exactly as the sequence of diagrams ε1, ε2, ε3,... ∈ TLN given by ε1 = ∩, ε2 = | ∩, ∪ ε3 = || ∩, and so on, with the parameter being the index, N = [A2,A1]. 0 (2) Since the orthogonal projection e1 : A1 → A0 commutes with A0 we have e1 ∈ P2, and by translation we obtain e1, . . . , ek−1 ∈ Pk for any k, and so TLN ⊂ P . The point now is that the planar algebra structure of TLN , obtained by composing diagrams, can be shown to extend into an abstract planar algebra structure of P . (3) This is something quite surprising, which follows from (1), via some clever positivity considerations, involving the Perron-Frobenius theorem. In fact, the subfactors having 2 π index N ∈ [1, 4] can be classiﬁed by ADE diagrams, and the obstruction N = 4 cos ( n ) itself comes from the fact that N must be the squared norm of such a graph. As it was the case with Theorem 11.1 above, our explanations here were very brief. For all this, and more, we recommend Jones’ papers [59], [60], [61], and [92].

Getting back now to the commuting squares, the idea is that any such square C produces a subfactor of the hyperﬁnite II1 factor R. Indeed, under suitable assumptions on the inclusions C00 ⊂ C10,C01 ⊂ C11, we can perform the basic construction for them, in ﬁnite dimensions, and we obtain a whole array of commuting squares, as follows:

A0 A1 A2 O O O

C02 / C12 / C22 / B2 O O O

C01 / C11 / C21 / B1 O O O

C00 / C10 / C20 / B0 Here the various A, B letters stand for the von Neumann algebras obtained in the limit, which are all isomorphic to the hyperﬁnite II1 factor R, and we have: Theorem 11.6. In the context of the above diagram, the following happen:

(1) A0 ⊂ A1 is a subfactor, and {Ai} is the Jones tower for it. 0 0 (2) The corresponding planar algebra is given by A0 ∩ Ak = C01 ∩ Ck0. (3) A similar result holds for the “horizontal” subfactor B0 ⊂ B1. Proof. Here (1) is something quite routine, (2) is a subtle result, called Ocneanu com- pactness theorem [74], and (3) follows from (1,2), by ﬂipping the diagram. HADAMARD MATRICES 171

Getting back now to the Hadamard matrices, we can extend our lineup of results, namely Theorem 11.3 and Theorem 11.4, with an advanced statement, as follows:

Theorem 11.7. Given a complex Hadamard matrix H ∈ MN (C), the diagram formed by the associated pair of orthogonal MASA, namely

∆ / MN (C) O O

C / H∆H∗ is a commuting square in the sense of subfactor theory, and the planar algebra P = (Pk) of the corresponding subfactor can be explicitely computed in terms of H. Proof. The fact that we have a commuting square is from Theorem 11.4, and the computation of the planar algebra is possible thanks to formula in Theorem 11.6. As for the precise formula of the planar algebra, which is something quite complicated, this can be found in [61], and we will be back to it later, in Theorem 11.9 below. Let us discuss now the relation with the quantum groups. We will need the following result, valid in the general context of the Hopf image construction:

Theorem 11.8. Given a matrix model π : C(G) → MK (C(T )), the fundamental corepresentation v of its Hopf is subject to the Tannakian conditions Hom(v⊗k, v⊗l) = Hom(U ⊗k,U ⊗l) where Uij = π(uij), and where the spaces on the right are taken in a formal sense. Proof. Since the morphisms increase the intertwining spaces, when deﬁned either in a representation theory sense, or just formally, we have inclusions as follows: Hom(u⊗k, u⊗l) ⊂ Hom(U ⊗k,U ⊗l) More generally, we have such inclusions when replacing (G, u) with any pair producing a factorization of π. Thus, by Tannakian duality [100], the Hopf image must be given by the fact that the intertwining spaces must be the biggest, subject to these inclusions. On the other hand, since u is biunitary, so is U, and it follows that the spaces on the right form a Tannakian category. Thus, we have a quantum group (H, v) given by: Hom(v⊗k, v⊗l) = Hom(U ⊗k,U ⊗l)

By the above discussion, C(H) follows to be the Hopf image of π, as claimed. With the above result in hand, we can compute the Tannakian category of the Hopf image, in the Hadamard matrix case, and we are led in this way to: 172 TEO BANICA

+ Theorem 11.9. The Tannakian category of the quantum group G ⊂ SN associated to a complex Hadamard matrix H ∈ MN (C) is given by T ∈ Hom(u⊗k, u⊗l) ⇐⇒ T ◦Gk+2 = Gl+2T ◦ where the objects on the right are constructed as follows: (1) T ◦ = id ⊗ T ⊗ id. jb P ¯ ¯ (2) Gia = k HikHjkHakHbk. (3) Gk = Gjkjk−1 ...Gj2j1 . i1...ik,j1...jk ikik−1 i2i1 Proof. With the notations in Theorem 11.8, we have the following formula: Hom(u⊗k, u⊗l) = Hom(U ⊗k,U ⊗l) The vector space on the right consists by deﬁnition of the complex N l × N k matrices T , satisfying the following relation: TU ⊗k = U ⊗lT If we denote this equality by L = R, the left term L is given by:

⊗k Lij = (TU )ij X ⊗k = TiaUaj a X = TiaUa1j1 ...Uakjk a As for the right term R, this is given by:

⊗l Rij = (U T )ij X ⊗l = Uib Tbj b X = Ui1b1 ...Uilbl Tbj b

Consider now the vectors ξij = Hi/Hj. Since these vectors span the ambient Hilbert space, the equality L = R is equivalent to the following equality:

< Lijξpq, ξrs >=< Rijξpq, ξrs > We use now the following well-known formula, expressing a product of rank one projections P1,...,Pk in terms of the corresponding image vectors ξ1, . . . , ξk:

< P1 ...Pkx, y >=< x, ξk >< ξk, ξk−1 > ...... < ξ2, ξ1 >< ξ1, y > HADAMARD MATRICES 173

This gives the following formula for L: X < Lijξpq, ξrs > = Tia < Pa1j1 ...Pakjk ξpq, ξrs > a X = Tia < ξpq, ξakjk > . . . < ξa1j1 , ξrs > a X = T Gqjk Gjkjk−1 ...Gj2j1 Gj1s ia pak akak−1 a2a1 a1r a X k+2 = TiaGrap,sjq a ◦ k+2 = (T G )rip,sjq As for the right term R, this is given by: X < Rijξpq, ξrs > = < Pi1b1 ...Pilbl ξpq, ξrs > Tbj b X = < ξpq, ξilbl > . . . < ξi1b1 , ξrs > Tbj b X b b = Gqbl G l l−1 ...Gb2b1 Gb1sT pil ilil−1 i2i1 i1r bj b X l+2 = Grip,sbqTbj b l+2 ◦ = (G T )rip,sjq Thus, we obtain the formula in the statement. See [13].

The point now is that, with k = 0, we obtain in this way precisely the spaces Pl computed by Jones in [61]. Thus, we are led to the following result:

Theorem 11.10. Let H ∈ MN (C) be a complex Hadamard matrix. ⊗k + (1) The planar algebra associated to H is given by Pk = F ix(u ), where G ⊂ SN is the associated quantum permutation group. P k (2) The corresponding Poincar´eseries f(z) = k dim(Pk)z equals the Stieltjes trans- R 1 P form G 1−zχ of the law of the main character χ = i uii. Proof. This follows by comparing the quantum group and subfactor results: (1) As already mentioned above, this simply follows by comparing Theorem 11.9 with the subfactor computation in [61]. For full details here, we refer to [13]. (2) This is a consequence of (1), and of the Peter-Weyl type results from [99], which tell us that ﬁxed points can be counted by integrating characters. Regarding now the subfactor itself, the result here is as follows: 174 TEO BANICA

Theorem 11.11. The subfactor associated to H ∈ MN (C) is of the form G N G A ⊂ (C ⊗ A) + with A = R o Gb, where G ⊂ SN is the associated quantum permutation group. Proof. This is something more technical, the idea being that the basic construction procedure for the commuting squares, explained before Theorem 11.6, can be performed in an “equivariant setting”, for commuting squares having components as follows:

G D ⊗G E = (D ⊗ (E o Gb)) To be more precise, starting with a commuting square formed by such algebras, we obtain by basic construction a whole array of commuting squares as follows, with {Di}, {Ei} being by deﬁnition Jones towers, and with D∞,E∞ being their inductive limits:

D0 ⊗G E∞ D1 ⊗G E∞ D2 ⊗G E∞ O O O

D0 ⊗G E2 / D1 ⊗G E2 / D2 ⊗G E2 / D∞ ⊗G E2 O O O

D0 ⊗G E1 / D1 ⊗G E1 / D2 ⊗G E1 / D∞ ⊗G E1 O O O

D0 ⊗G E0 / D1 ⊗G E0 / D2 ⊗G E0 / D∞ ⊗G E0 The point now is that this quantum group picture works in fact for any commuting square having C in the lower left corner. In the Hadamard matrix case, that we are interested in here, the corresponding commuting square is as follows:

N N N C ⊗G C / C ⊗G C O O

N C ⊗G C / C ⊗G C Thus, the subfactor obtained by vertical basic construction appears as follows:

N C ⊗G E∞ ⊂ C ⊗G E∞

But this gives the conclusion in the statement, with the II1 factor appearing there being by deﬁnition A = E∞ o Gb, and with the remark that we have E∞ ' R. See [4]. HADAMARD MATRICES 175

All this is of course quite heavy, with the above results being subject to several extensions, and with all this involving several general correspondences between quantum groups, planar algebras, commuting squares and subfactors, that we will not get into. As a technical comment here, it is possible to deduce Theorem 11.10 directly from Theorem 11.11, via some quantum group computations. However, Theorem 11.11 and its proof involve some heavy algebra and functional analysis, coming on top of the heavy algebra and functional analysis required for the general theory of the commuting squares, and this makes the whole thing quite unusable, in practice. Thus, while being technically weaker, Theorem 11.10 above remains the main result on the subject. We refer to [4], [13], [21] and related papers for the full story of all this.

As already mentioned in the beginning of this section, all this is conjecturally related to statistical mechanics. Indeed, the Tannakian category/planar algebra formula from Theorem 11.9 has many similarities with the transfer matrix computations for the spin models, and this is explained in Jones’ paper [61], and known for long before that, from his 1989 paper [60]. However, the precise significance of the Hadamard matrices in statistical mechanics, or in related areas such as link invariants, remains a bit unclear. From a quantum group perspective, the same questions make sense. The idea here, which is old folklore, going back to the 1998 discovery by Wang [96] of the quantum + permutation group SN , is that associated to any 2D spin model should be a quantum + + permutation group G ⊂ SN , which appears by factorizing the flat representation C(SN ) → MN (C) associated to the N ×N matrix of the Boltzmann weights of the model, and whose representation theory computes the partition function of the model. This is supported on one hand by Jones’ theory in [60], [61], via the connecting results presented above, and on the other hand by a number of more recent results, such as those in [16], having similarities with the computations for the Ising and Potts models. However, the whole thing remains not axiomatized, at least for the moment, and in what regards the Hadamard matrices, their precise physical significance remains unclear.

Getting back to work now, the above discussion suggests heavily investing time and energy into the computation of integrals over Hopf images, because it is via such integrals that the mathematics of the corresponding lattice model is supposed to appear. To be more precise, we would like for instance to have advanced representation theory results, of probabilistic ﬂavor, in the spirit of [36], [45], [46].

Let us begin with some generalities. Our claim is that the “good” problem, about any compact quantum group, is that of computing the law of the main character. This claim, which is something well-known, and generally agreed upon, is supported by a wealth of interesting results, which can be summarized as follows: 176 TEO BANICA

Theorem 11.12. Given a Woronowicz algebra (A, u), the law of the main character

N X χ = uii i=1 with respect to the Haar integration has the following properties: ⊗k (1) The moments of χ are the numbers Mk = dim(F ix(u )). (2) Mk counts as well the lenght p loops at 1, on the Cayley graph of A. (3) law(χ) is the Kesten measure of the associated discrete quantum group. (4) When u ∼ u¯ the law of χ is a usual measure, supported on [−N,N]. (5) The algebra A is amenable precisely when N ∈ supp(law(Re(χ))). (6) Any morphism f :(A, u) → (B, v) must increase the numbers Mk. (7) Such a morphism f is an isomorphism when law(χu) = law(χv). Proof. All this is quite advanced, the idea being as follows: (1) This comes from the Peter-Weyl type theory in [99], which tells us the number of ⊗k k fixed points of v = u can be recovered by integrating the character χv = χu. ∗ (2) This is something true, and well-known, for A = C (Γ), with Γ =< g1, . . . , gN > being a discrete group. In general, the proof is quite similar. (3) This is actually the definition of the Kesten measure, in the case A = C∗(Γ), with Γ =< g1, . . . , gN > being a discrete group. In general, this follows from (2). ∗ (4) The equivalence u ∼ u¯ translates into χu = χu, and this gives the first assertion. ∗ As for the support claim, this follows from uu = 1 =⇒ ||uii|| ≤ 1, for any i. (5) This is the Kesten amenability criterion, which can be established as in the classical ∗ case, A = C (Γ), with Γ =< g1, . . . , gN > being a discrete group. (6) This is something elementary, which follows from (1) above, and from the fact that the morphisms of Woronowicz algebras increase the spaces of fixed points. (7) This follows by using (6), and the Peter-Weyl type theory from [99], the idea being ⊗k that if f is not injective, then it must strictly increase one of the spaces F ix(u ). As a conclusion, computing µ = law(χ) is indeed the main question to be solved, from a massive number of mathematical viewpoints. In addition to all this, in view of the above, the measure µ = law(χ) is expected to have an interesting physical meaning.

+ More concretely now, let us first investigate the quantum groups SN ,SN . For the symmetric group SN that the standard coordinates are given by uij = χ(σ|σ(j) = i), and so the main character counts the number of fixed points: n o X χ(σ) = δσ(i),i = # i ∈ {1,...,N} σ(i) = i i A well-known computation, based on the inclusion-exclusion principle, shows that with 1 N → ∞ the probability for a random permutation σ ∈ SN to have no fixed points is ' e . HADAMARD MATRICES 177

More generally, one can show that the probability for σ ∈ SN to have exactly k ﬁxed 1 points is ' k!e . Thus, µ = law(χ) becomes with N → ∞ a Poisson variable. + Regarding now SN , the computation here is a bit more complicated, leading to a “free version” of the Poisson law. In order to explain this, we will need the following result, with ∗ being the classical convolution, and being Voiculescu’s free convolution: Theorem 11.13. The following Poisson type limits converge, for any t > 0, 1 1 ∗n 1 1 n pt = lim 1 − δ0 + δt , πt = lim 1 − δ0 + δt n→∞ n n n→∞ n n the limiting measures being the Poisson law pt, and the Marchenko-Pastur law πt, ∞ k p 2 1 X t δk 4t − (x − 1 − t) p = , π = max(1 − t, 0)δ + dx t et k! t 0 2πx k=0 whose moments are given by the following formulae:

X |π| X |π| Mk(pt) = t ,Mk(πt) = t π∈P (k) π∈NC(k)

The Marchenko-Pastur measure πt is also called free Poisson law. Proof. This is something quite advanced, related to probability theory, free probability theory, and random matrices, the idea being as follows: (1) The ﬁrst step is that of ﬁnding suitable functional transforms, which linearize the convolution operations in the statement. In the classical case this is the logarithm of the Fourier transform log F , and in the free case this is Voiculescu’s R-transform. (2) With these tools in hand, the above limiting theorems can be proved in a standard way, a bit as when proving the Central Limit Theorem. The computations give the moment formulae in the statement, and the density computations are standard as well. (3) Finally, in order for the discussion to be complete, what still remains to be explained is the precise nature of the “liberation” operation pt → πt, as well as the random matrix occurrence of πt. This is more technical, and we refer here to [27], [67], [95]. Getting back now to quantum groups, the results here are as follows: P Theorem 11.14. The law of χ = i uii is as follows: (1) For SN with N → ∞ we obtain the Poisson law p1. + (2) For SN with N ≥ 4 we obtain the free Poisson law π1. P[tN] + Also, the law of χt = i=1 uii for SN /SN , with t ∈ (0, 1], becomes pt/πt with N → ∞. Proof. This is something quite technical, the idea being as follows: (1) In the classical case this is well-known, and follows for instance by using the inclusion-exclusion principle, and then letting N → ∞, as explained above. 178 TEO BANICA

⊗k (2) In the free case it is known that Pk = F ix(u ) equals TLN (k) at N ≥ 4, and at the probabilistic level, this leads to the formulae in the statement. See [12]. Let us go back now to the Hadamard matrices, and do some computations here. In the general matrix model context, from Deﬁnition 10.9 above, we have the following formula for the Haar integration functional of the Hopf image, coming from [99]:

Theorem 11.15. Given an inner faithful model π : C(G) → MK (C(T )), we have k Z 1 X Z r = lim k→∞ k G r=1 G R r ∗r R where G = (ϕ ◦ π) , with ϕ = tr ⊗ T being the random matrix trace. R 0 Proof. We must prove that the limit in the statement G converges, and that we have R 0 R G = G. It is enough to check this on the coeﬃcients of corepresentations: Z 0 Z id ⊗ v = id ⊗ v G G We know from Theorem 10.8 that the matrix on the right is the orthogonal projection onto F ix(v). As for the matrix on the left, this is the orthogonal projection onto the 1-eigenspace of (id ⊗ ϕπ)v. Now observe that, if we set Vij = π(vij), we have: (id ⊗ ϕπ)v = (id ⊗ ϕ)V Thus, as in the proof of Theorem 10.8, we conclude that the 1-eigenspace that we are interested in equals F ix(V ). But, according to Theorem 11.8, we have: F ix(V ) = F ix(v) R 0 R Thus, we have proved that we have G = G, as desired. R r In practice, we are led to the computation of the truncated integrals G appearing in the above result, and the formula of these truncated integrals is as follows: R r ∗r Proposition 11.16. The truncated integrals G = (ϕ ◦ π) are given by Z r uε1 . . . uεp = (T r) a1b1 apbp ε a1...ap,b1...bp G for any exponents εi ∈ {1, ∗}, with the matrix on the right being given by Z (T ) = tr ⊗ (U ε1 ...U εp ) ε i1...ip,j1...jp i1j1 ipjp T where Uij = π(uij) are the images of the standard coordinates in the model. Proof. This is something straightforward, which comes from the deﬁnition of the truncated R r ∗r R integrals, namely G = (ϕ ◦ π) , with ϕ = tr ⊗ T being the random matrix trace. HADAMARD MATRICES 179

Regarding now the main character, the result here is as follows:

r R r ∗r Theorem 11.17. Let µ be the law of χ = T r(u) with respect to G = (ϕ ◦ π) . 1 Pk r (1) We have the convergence formula µ = limk→∞ k r=0 µ , in moments. r r r (2) The ∗-moments of the truncated measure µ are the numbers cε = T r(Tε ). Proof. These results are both elementary, the proof being as follows: (1) This follows from the general limiting formula in Theorem 11.15. (2) This follows from the formula in Proposition 11.16, by summing over ai = bi. In connection with the complex Hadamard matrices, we can use this technology in order to discuss the behavior of the construction H → G with respect to the operations H → Ht, H,H¯ ∗. Let us ﬁrst introduce the following abstract duality:

Definition 11.18. Let π : C(G) → MN (C) be inner faithful, mapping uij → Uij. 0 + 0 (1) We set (Ukl)ij = (Uij)kl, and define ρ : C(UN ) → MN (C) by vkl → Ukl. e 0 (2) We perform the Hopf image construction, as to get a model ρ : C(G ) → MN (C). + In this definition UN is Wang’s quantum unitary group, whose standard coordinates are subject to the biunitarity condition u∗ = u−1, ut =u ¯−1. Observe that the matrix U 0 constructed in (1) is given by U 0 = ΣU, where Σ is the flip. Thus this matrix is indeed biunitary, and produces a representation ρ as in (1), and then a factorization as in (2). The operation A → A0 is a duality, in the sense that we have A00 = A, and in the Hadamard matrix case, this comes from the operation H → Ht. See [11]. We denote by D the dilation operation for probability measures, or for general ∗- distributions, given by the formula Dr(law(X)) = law(rX). We have then:

r r Theorem 11.19. Consider the rescaled measure η = D1/N (µ ). r r p r r p 0 (1) The moments γp = cp/N of η satisfy γp(G) = γr (G ). r 0 0 (2) η has the same moments as the matrix Tr = Tr(G ). r 0 (3) In the real case u =u ¯ we have η = law(Tr). Proof. All the results follow from Theorem 11.17, as follows: (1) We have the following computation:

r X c (A) = (T ) 1 1 2 2 ...... (T ) r r 1 1 p p i1...ip,i1...ip p i1...ip,i1...ip i X = tr(U 1 2 ...U 1 2 ) ...... tr(U r 1 ...U r 1 ) i1i1 ipip i1i1 ipip i 1 X X = (Ui1i2 )j1j1 ... (Ui1i2 )j1j1 ...... (Uiri1 )jrjr ... (Uir i1 )jrjr N r 1 1 1 2 p p p 1 1 1 1 2 p p p 1 i j 180 TEO BANICA

0 In terms of the matrix (Ukl)ij = (Uij)kl, then by permuting the terms in the product b a b a on the right, and ﬁnally with the changes ia ↔ ib , ja ↔ jb , we obtain:

r 1 X X 0 0 0 0 cp(A) = (Uj1j1 )i1i2 ... (Uj1j1 )i1i2 ...... (Ujrjr )iri1 ... (Ujrjr )ir i1 N r 1 2 1 1 p 1 p p 1 2 1 1 p 1 p p i j

1 X X 0 0 0 0 = (Uj1j1 )i1i2 ... (Ujrjr )iri1 ...... (Uj1j1 )i1i2 ... (Ujrjr )ir i1 N r 1 2 1 1 1 2 1 1 p 1 p p p 1 p p i j

1 X X 0 0 0 0 1 1 1 1 p p p p p p = (Uj1j2 )i i ... (Uj1j2 )i i ...... (Uj j1 )i i ... (Uj j1 )ir i N r 1 1 1 2 r r r 1 1 1 1 2 r r 1 i j r On the other hand, if we use again the above formula of cp(A), but this time for the matrix U 0, and with the changes r ↔ p and i ↔ j, we obtain:

p 0 1 X X 0 0 0 0 1 1 1 1 p p p p p p cr(A ) = (Uj1j2 )i i ... (Uj1j2 )i i ...... (Uj j1 )i i ... (Uj j1 )ir i N p 1 1 1 2 r r r 1 1 1 1 2 r r 1 i j r r p p 0 Now by comparing this with the previous formula, we obtain N cp(A) = N cr(A ). r p p 0 r Thus we have cp(A)/N = cr(A )/N , and this gives the result. (2) By using (1) and the formula in Theorem 11.17, we obtain: cr(A) cp(A0) T r((T 0)p) p = r = r = tr((T 0)p) N p N r N r r But this gives the equality of moments in the statement. (3) This follows from the moment equality in (2), and from the standard fact that for self-adjoint variables, the moments uniquely determine the distribution. All this is interesting in connection with the transposition operation H → Ht for the complex Hadamard matrices, and its relation with the lattice model problematics. Indeed, in the context of the classical spin models, the matrix of Boltzmann weights must be symmetric, and the precise meaning of the “Hadamard matrix models” depends on this. For more on this, and other related issues, we refer to [11] and related papers. HADAMARD MATRICES 181

12. Fourier models

We have seen that associated to any Hadamard matrix H ∈ MN (C) is a quantum + permutation group G ⊂ SN . The construction H → G is something very simple, obtained + by factorizing the representation π : C(SN ) → MN (C) given by uij → P roj(Hi/Hj), N where H1,...,HN ∈ T are the rows of H. As a basic example, a Fourier matrix H = FG produces in this way the group G itself, acting on itself.

+ Generally speaking, the quantum group G ⊂ SN is expected to capture the mathematics and physics of the Hadamard matrix H ∈ MN (C), via its representation theory. All this is, unfortunately, a bit conjectural for the moment. In what concerns von Neumann algebra, orthogonal MASA, commuting square, subfactor and planar algebra aspects, this is deﬁnitely the case, thanks to the results explained in section 11. Getting beyond this level, however, as to reach to some clear statistical mechanical results, in the spirit of [59], [60], [61] and beyond, remains an open problem.

From a purely mathematical perspective, there are many interesting questions which are open as well. We would like for instance to know if the defect, and the other geometric invariants of H, are captured or not by the representation theory of G. Similar questions make sense for the glow. Nothing much is known here, and for some comments and speculations on this subject, we refer to [4], [5], [6], [7], [8], [9].

In the lack of an answer to these questions, let us go back to the construction H → G, as it is, and try to have more examples worked out. Generally speaking, going beyond Theorem 10.16 is a diﬃcult task, and only one computation is available so far. This computation, performed in [33] by using the subfactor formalism, and in [11] by using the quantum group formalism, regards the Dit¸˘adeformations of the tensor products FG×H = FG ⊗ FH of Fourier matrices. Besides [11], [33], some further results on all this are available from [9] and from [28]. We will follow here the approach in [11].

Before starting, we should mention that, in view of the results from section 4 above, the natural question regarding the deformed Fourier matrices would be that of computing the quantum groups associated to the Nicoara-White deformations of FG. However, no result is available here so far, the point being that the above-mentioned papers [9], [11], [28], [33] were all written before the Nicoara-White discovery in [73]. In short, we have to be modest here as well. In what follows we will explain the material from the paper [11], which is somehow central to the subject. This will consist in the computation for the Dit¸˘adeformations of the tensor products FG×H = FG ⊗FH , at generic values of the parameters, and of some related probabilistic work.

Let us begin by recalling, following [47], the deﬁnition of the deformations: 182 TEO BANICA

G×H Proposition 12.1. The matrix FG×H ∈ MG×H (T ) given by

(FG×H )ia,jb(Q) = Qib(FG)ij(FH )ab is complex Hadamard, and its ﬁber at Q = (1ib) is the Fourier matrix FG×H .

Proof. The fact that the rows of FG ⊗Q FH = FG×H (Q) are pairwise orthogonal follows from definitions. With 1 = (1ij) we have (FG ⊗1 FH )ia,jb = (FG)ij(FH )ab, and we recognize here the formula of FG×H = FG ⊗ FH , in double index notation. As in [11], it is convenient to take an abstract approach to all this: Definition 12.2. Associated to a finite abelian group X is the Fourier model

π : C(X) → M|X|(C) 1 coming from the matrix (Uij)kl = N Fi−j,k−l, where F = FX . Now let X,Y be finite abelian groups, and let us try to understand the model constructed by deforming the tensor product of the corresponded Fourier models: Definition 12.3. Given two finite abelian groups X,Y , we consider the corresponding Fourier models U, V , we construct the deformation W = U ⊗Q V , and we factorize π + Q C(SX×Y ) / MX×Y (C) 9 π % C(GQ) with C(GQ) being the Hopf image of πQ.

Explicitely computing the compact quantum group GQ, as function of the parameter matrix Q ∈ MX×Y (T), will be our main purpose, in what follows. In order to do so, we use the following notion: + + Deﬁnition 12.4. Let C(SM ) → A and C(SN ) → B be Hopf algebra quotients, with fundamental corepresentations denoted u, v. We let ∗N (i) A ∗w B = A ∗ B/ < [uab , vij] = 0 > (i) with the Hopf algebra structure making wia,jb = uab vij a corepresentation. The fact that we have indeed a Hopf algebra follows from the fact that w is magic. In terms of quantum groups, if A = C(G), B = C(H), we write A ∗w B = C(G o∗ H):

C(G) ∗w C(H) = C(G o∗ H)

The o∗ operation is then the free analogue of o, the usual wreath product. See [11]. We will need as well the following elementary result: HADAMARD MATRICES 183

Proposition 12.5. If X is a ﬁnite abelian group then + C(X) = C(SX )/ < uij = ukl|∀i − j = k − l > with all the indices taken inside X. + Proof. Observe ﬁrst that C(Y ) = C(SX )/ < uij = ukl|∀i − j = k − l > is commutative, because uijukl = uijui,l−k+i = δj,l−k+iuij and ukluij = ui,l−k+iuij = δj,l−k+iuij. Thus we have Y ⊂ SX , and since uij(σ) = δiσ(j) for any σ ∈ Y , we obtain: i − j = k − l =⇒ (σ(j) = i ⇐⇒ σ(l) = k) But this condition tells us precisely that σ(i) − i must be independent on i, and so σ(i) = i + x for some x ∈ X, and so σ ∈ X, as desired.

We can now factorize representation πQ in Deﬁnition 12.3, as follows: Theorem 12.6. We have a factorization π + Q C(SX×Y ) / MX×Y (C) 8 π & C(Y o∗ X) (i) P P given by Uab = j Wia,jb and by Vij = a Wia,jb, independently of b.

Proof. With K = FX ,L = FY and M = |X|,N = |Y |, the formula of the magic matrix W ∈ MX×Y (MX×Y (C)) associated to H = K ⊗Q L is:

1 QicQjd KikKjl LacLbd (Wia,jb)kc,ld = · · · MN QidQjc KilKjk LadLbc

1 QicQjd = · · Ki−j,k−lLa−b,c−d MN QidQjc

Our claim that the representation πQ constructed in Deﬁnition 12.3 can be factorized in three steps, up to the factorization in the statement, as follows: π + Q C(SX×Y ) / MX×Y (C) 4 9 O

+ + + C(SY o∗ SX ) / C(SY o∗ X) / C(Y o∗ X) Indeed, the construction of the map on the left is standard, and this produces the ﬁrst factorization. Regarding the second factorization, this comes from the fact that since the elements Vij depend on i − j, they satisfy the deﬁning relations for the quotient algebra + C(SX ) → C(X), coming from Proposition 12.5. Finally, regarding the third factorization, 184 TEO BANICA observe that the above matrix Wia,jb depends only on a−b. By summing over j we obtain (i) that Uab depends only on a − b, and we are done. In order to further factorize the above representation, we use:

Deﬁnition 12.7. If H y Γ is a ﬁnite group acting by automorphisms on a discrete group, the corresponding crossed coproduct Hopf algebra is ∗ ∗ C (Γ) o C(H) = C (Γ) ⊗ C(H) with comultiplication given by the following formula,

X −1 ∆(r ⊗ δk) = (r ⊗ δh) ⊗ (h · r ⊗ δh−1k) h∈H for r ∈ Γ and k ∈ H. Observe that C(H) is a subcoalgebra, and that C∗(Γ) is not a subcoalgebra. The quantum group corresponding to C∗(Γ) o C(H) is denoted Γb o H. Now back to the factorization in Theorem 12.6, the point is that we have:

Proposition 12.8. With L = FY ,N = |Y | we have an isomorphism ∗ ∗X C(Y o∗ X) ' C (Y ) o C(X) (i) 1 P (i) given by vij → 1 ⊗ vij and uab = N c Lb−a,cc ⊗ 1. ∗X Proof. We know that C(Y o∗ X) is the quotient of C(Y ) ∗ C(X) by the relations (i) [uab , vij] = 0. Now since vij depends only on j − i, we obtain: (i) (i) [uab , vkl] = [uab , vi,l−k+i] = 0 Thus, we are in a usual tensor product situation, and we have: ∗X C(Y o∗ X) = C(Y ) ⊗ C(X) Let us compose now this identiﬁcation with Φ∗X ⊗ id, where Φ : C(Y ) → C∗(Y ) is the Fourier transform. We obtain an isomorphism as in the statement, and since Φ(uab) = 1 P (i) N c Lb−a,cc, the formula for the image of uab is indeed the one in the statement. Here is now our key result, which will lead to further factorizations:

(i) P (i) P Proposition 12.9. With c = a Lacua0 and εke = i Kikeie we have:

(i) Qi,e−cQi−k,e π(c )(εke) = εk,e−c QieQi−k,e−c

(i1) (is) In particular if c1 + ... + cs = 0 then π(c1 . . . cs ) is diagonal, for any i1, . . . , is. HADAMARD MATRICES 185

Proof. We have the following formula: (i) X (i) X π(c ) = Lacπ(ua0 ) = LacWia,j0 a aj On the other hand, in terms of the basis in the statement, we have:

1 X QidQje Wia,jb(εke) = δi−j,k La−b,d−eεkd N QieQjd d We therefore obtain, as desired:

(i) 1 X QidQi−k,e π(c )(εke) = Lac La,d−eεkd N QieQi−k,d ad 1 X QidQi−k,e X = εkd La,d−e+c N QieQi−k,d d a X QidQi−k,e = εkdδd,e−c QieQi−k,d d Qi,e−cQi−k,e = εk,e−c QieQi−k,e−c Regarding now the last assertion, this follows from the fact that each matrix of type (ir) π(cr ) acts on the standard basis elements εke by preserving the left index k, and by rotating by cr the right index e. Thus when we assume c1 +...+cs = 0 all these rotations compose up to the identity, and we obtain indeed a diagonal matrix. We have now all needed ingredients for reﬁning Theorem 12.6, as follows: Theorem 12.10. We have a factorization as follows, π + Q C(SX×Y ) / MX×Y (C) 6 ρ ( ∗ C (ΓX,Y ) o C(X)

∗X (i1) (is) (j1) (js) P P where ΓX,Y = Y / < [c1 . . . cs , d1 . . . ds ] = 1| r cr = r dr = 0 >. ∗ Proof. Assume that we have a representation π : C (Γ) o C(X) → ML(C), let Λ be a X-stable normal subgroup of Γ, so that X acts on Γ/Λ and that we can form the crossed coproduct C∗(Γ/Λ) o C(X), and assume that π is trivial on Λ. Then π factorizes as: ∗ π C (Γ) o C(X) / ML(C) 6 ρ ) C∗(Γ/Λ) o C(X) 186 TEO BANICA

∗X With Γ = Y , and by using the above results, this gives the result.

In general, further factorizing the representation found in Theorem 12.10 is a quite complicated task. In what follows we restrict attention to the case where the parameter matrix Q is generic, in the sense that its entries are as algebrically independent as possible, and we prove that the representation in Theorem 12.10 is the minimal one. Our starting point is the group ΓX,Y found above: Definition 12.11. Associated to two finite abelian groups X,Y is the discrete group * + . (i ) (j ) X X ∗X 1 (is) 1 (js) ΓX,Y = Y [c1 . . . cs , d1 . . . ds ] = 1 cr = dr = 0 r r where the superscripts refer to the X copies of Y , inside the free product. We will need a more convenient description of this group. The idea here is that the above commutation relations can be realized inside a suitable semidirect product. Given a group acting on another group, H y G, we denote as usual by G o H the semidirect product of G by H, which is the set G × H, with multiplication: (a, s)(b, t) = (as(b), st) Now given a group G, and a finite abelian group Y , we can make Y act on GY , and form the product GY o Y . Since the elements of type (g, . . . , g) are invariant, we can form as well the prod- Y Y |Y |−1 uct (G /G) o Y , and by identifying G /G ' G via the map (1, g1, . . . , g|Y |−1) → |Y |−1 (g1, . . . , g|Y |−1), we obtain a product G o Y . With these notations, we have the following result:

Proposition 12.12. The group ΓX,Y has the following properties: (|X|−1)(|Y |−1) (1) ΓX,Y ' Z o Y . (|X|−1)|Y | (0) (i) (2) ΓX,Y ⊂ Z o Y via c → (0, c) and c → (bi0 − bic, c) for i 6= 0, where bic are the standard generators of Z(|X|−1)|Y |. Proof. We prove these assertions at the same time. We must prove that we have group morphisms, given by the formulae in the statement, as follows: (|X|−1)(|Y |−1) (|X|−1)|Y | ΓX,Y ' Z o Y ⊂ Z o Y Our ﬁrst claim is that the formula in (2) deﬁnes a morphism as follows:

(|X|−1)|Y | ΓX,Y → Z o Y Indeed, the elements (0, c) produce a copy of Y , and since we have a group embedding |Y | (i) Y ⊂ Z o Y given by c → (b0 − bc, c), the elements C = (bi0 − bic, c) produce a copy HADAMARD MATRICES 187 of Y , for any i 6= 0. In order to check now the commutation relations, observe that: ! (i1) (is) X C1 ...Cs = bi10 − bi1c1 + bi2c1 − bi2,c1+c2 + ... + bis,c1+...+cs−1 − bis,c1+...+cs , cr r

P (i1) (is) (|X|−1)|Y | Thus r cr = 0 implies C1 ...Cs ∈ Z , and since we are now inside an abelian group, we have the commutation relations, and our claim is proved. Using the considerations before the statement of the proposition, it is routine to construct an embedding Z(|X|−1)(|Y |−1) o Y ⊂ Z(|X|−1)|Y | o Y such that we have group morphisms whose composition is the group morphism just constructed, as follows: (|X|−1)(|Y |−1) (|X|−1)|Y | ΓX,Y → Z o Y ⊂ Z o Y It remains to prove that the map on the left is injective. For this purpose, consider the (i) morphism ΓX,Y → Y given by c → c, whose kernel T is formed by the elements of type (i1) (is) P c1 . . . cs , with r cr = 0. We get an exact sequence, as follows:

1 → T → ΓX,Y → Y → 1 (0) This sequence splits by c → c , so we have ΓX,Y ' T o Y . Now by the deﬁnition of ΓX,Y , the subgroup T constructed above is abelian, and is moreover generated by the elements (−c)(0)c(i), i, c 6= 0. Finally, the fact that T is freely generated by these elements follows from the computation in the proof of Proposition 12.14 below. Let us specify now what our genericity assumptions are: Deﬁnition 12.13. We use the following notions:

(1) We call p1, . . . , pm ∈ T root independent if for any r1, . . . , rm ∈ Z we have r1 rm p1 . . . pm = 1 =⇒ r1 = ... = rm = 0. (2) A matrix Q ∈ MX×Y (T), taken to be dephased (Q0c = Qi0 = 1), is called generic if the elements Qic, with i, c 6= 0, are root independent. We will need the following technical result:

Proposition 12.14. Assume that Q ∈ MX×Y (T) is generic, and put

ke Qi,e−cQi−k,e θic = QieQi−k,e−c k For every k ∈ X, we have a representation π :ΓX,Y → U|Y | given by:

k (i) ke π (c )e = θic e−c k The family of representations (π )k∈X is projectively faithful in the sense that if for some k t ∈ ΓX,Y , we have that π (t) is a scalar matrix for any k, then t = 1. 188 TEO BANICA

k Proof. The representations π arise from above. With ΓX,Y = T o Y , as in the proof of k Proposition 12.12, we see that for t ∈ ΓX,Y such that π (t) is a scalar matrix for any k, then t ∈ T , since the elements of T are the only ones having their image by πk formed by diagonal matrices. Now write t as follows, with the generators of T being as in the proof of Proposition 12.12 above, and with Ric ∈ Z being certain integers: Y t = ((−c)(0)(c)(i))Ric i6=0,c6=0 Consider now the following quantities:

−1 Y Y ke ke Ric A(k, e) = (θic (θ0c ) ) i6=0 c6=0

Y Y ke Ric ke −Ric = (θic ) (θ0c ) i6=0 c6=0 P Y Y ke Ric Y ke − i6=0 Ric = (θic ) · (θ0c ) i6=0 c6=0 c6=0 P Y Y ke Rjc Y Y ke i6=0 Ric = (θjc ) · (θjc ) j6=0 c6=0 c6=0 j6=0 P Y Y ke Rjc+ i6=0 Ric = (θjc ) j6=0 c6=0

k We have π (t)(e) = A(k, e)e for any k, e. Our assumption is that for any k, we have A(k, e) = A(k, f) for any e, f. Using the root independence of the elements Qic, i, c 6= 0, we see that this implies Ric = 0 for any i, c, and this proves our assertion. We will need as well the following result, technical as well:

Proposition 12.15. Let π : C∗(Γ) o C(H) → L be a surjective Hopf algebra map, such that π|C(H) is injective, and such that for r ∈ Γ and f ∈ C(H), we have: π(r ⊗ 1) = π(1 ⊗ f) =⇒ r = 1 Then π is an isomorphism.

Proof. We use here various Hopf algebra tools. Put A = C∗(Γ) o C(H). We start with the following Hopf algebra exact sequence, where i(f) = 1 ⊗ f and p = ε ⊗ 1:

i p ∗ C → C(H) → A → C (Γ) → C Since π ◦ i is injective, and Hopf subalgebra π ◦ i(C(H)) is central in L, we can form the quotient Hopf algebra L = L/(π ◦ i(C(H))+L, and we get another exact sequence:

π◦i q C → C(H) −−→ L → L → C HADAMARD MATRICES 189

Note that this sequence is indeed exact, e.g. by centrality. So we get the following diagram with exact rows, with the Hopf algebra map on the right surjective:

i p C / C(H) / A / C∗(Γ) / C

π◦i q C / C(H) / L / L / C

Since a quotient of a group algebra is still a group algebra, we get a commutative diagram with exact rows as follows:

i p C / C(H) / A / C∗(Γ) / C

π◦i q0 C / C(H) / L / C∗(Γ) / C

Here the Hopf algebra map on the right is induced by a surjective morphism u :Γ → Γ, g 7→ g. By the ﬁve lemma we just have to show that u is injective. So, let g ∈ Γ be such that u(g) = 1. Then q0π(g ⊗ 1) = up(g ⊗ 1) = u(g) = g = 1. For g ∈ Γ, put:

gA = {a ∈ A | p(a1) ⊗ a2 = g ⊗ a}

0 gL = {l ∈ L | q (l1) ⊗ l2 = g ⊗ l}

The commutativity of the right square ensures that π(gA) ⊂ gL. Then with the previous g, we have π(g ⊗ 1) ∈ 1L = πi(C(H)) (exactness of the sequence), so π(g ⊗ 1) = π(1 ⊗ f) for some f ∈ C(H). We conclude by our assumption that g = 1.

We have now all the needed ingredients for proving a main result, as follows:

Theorem 12.16. When Q is generic, the minimal factorization for πQ is

π + Q C(SX×Y ) / MX×Y (C) 6 π ( ∗ C (ΓX,Y ) o C(X)

(|X|−1)(|Y |−1) where ΓX,Y ' Z o Y is the discrete group constructed above.

∗ Proof. We want to apply Proposition 12.14 to the morphism θ : C (ΓX,Y ) o C(X) → L arising from the factorization in Theorem 12.10, where L denotes the Hopf image of πQ, 190 TEO BANICA which produces the following commutative diagram: π + Q C(SX×Y ) / MX×Y (C) 6 @

( L O π θ % ∗ C (ΓX,Y ) o C(X) The ﬁrst observation is that the injectivity assumption on C(X) holds by construction, and that for f ∈ C(X), the matrix π(f) is “block scalar”, the blocks corresponding to the indices k in the basis εke in the basis from Proposition 12.14. Now for r ∈ ΓX,Y with θ(r ⊗ 1) = θ(1 ⊗ f) for some f ∈ C(X), we see, using the commutative diagram, that we will have that π(r ⊗ 1) is block scalar. k By Proposition 12.12, the family of representations (π ) of ΓX,Y , corresponding to the blocks k, is projectively faithful, so r = 1. We can apply indeed Proposition 12.14, and we are done. Let us try now to compute the Kesten measure µ = law(χ). Our results here will be a combinatorial moment formula, a geometric interpretation of it, and an asymptotic result. Let us begin with the moment formula, which is as follows: Theorem 12.17. We have the moment formula Z 1 p i1, . . . , ip ∈ X [(i1, d1), (i2, d2),..., (ip, dp)] χ = # G |X| · |Y | d1, . . . , dp ∈ Y = [(i1, dp), (i2, d1),..., (ip, dp−1)] where the sets between square brackets are by deﬁnition sets with repetition. Proof. According to the various formulae above, the factorization found in Theorem 12.16 is, at the level of standard generators, as follows: + ∗ C(SX×Y ) → C (ΓX,Y ) ⊗ C(X) → MX×Y (C) 1 P (i) uia,jb → |Y | c Fb−a,cc ⊗ vij → Wia,jb Thus, the main character is given by: ! 1 X X X χ = c(i) ⊗ v = c(i) ⊗ v = c(i) ⊗ δ |Y | ii ii 1 iac ic ic Now since the Haar functional of C∗(Γ) o C(H) is the tensor product of the Haar functionals of C∗(Γ),C(H), this gives the following formula, valid for any p ≥ 1: !p Z 1 Z X χp = c(i) |X| G ΓbX,Y ic HADAMARD MATRICES 191

P (i) Let Si = c . By using the embedding in Proposition 12.12 (2), with the notations c P there we have Si = c(bi0 − bic, c), and these elements multiply as follows:

 b − b + b − b  X i10 i1c1 i2c1 i2,c1+c2 Si1 ...Sip =  +bi3,c1+c2 − bi3,c1+c2+c3 + ...... , c1 + ... + cp 

c1...cp ...... + bip,c1+...+cp−1 − bip,c1+...+cp

In terms of the new indices dr = c1 + ... + cr, this formula becomes:

b − b + b − b  X i10 i1d1 i2d1 i2d2 Si1 ...Sip =  +bi3d2 − bi3d3 + ...... , dp 

d1...dp ...... + bipdp−1 − bipdp

Now by integrating, we must have dp = 0 on one hand, and on the other hand:

[(i1, 0), (i2, d1),..., (ip, dp−1)] = [(i1, d1), (i2, d2),..., (ip, dp)]

Equivalently, we must have dp = 0 on one hand, and on the other hand:

[(i1, dp), (i2, d1),..., (ip, dp−1)] = [(i1, d1), (i2, d2),..., (ip, dp)]

Thus, by translation invariance with respect to dp, we obtain:

Z 1 [(i1, d1), (i2, d2),..., (ip, dp)] Si1 ...Sip = # d1, . . . , dp ∈ Y = [(i1, dp), (i2, d1),..., (ip, dp−1)] ΓbX,Y |Y |

It follows that we have the following moment formula:

Z !p X 1 i1, . . . , ip ∈ X [(i1, d1), (i2, d2),..., (ip, dp)] Si = # |Y | d1, . . . , dp ∈ Y = [(i1, dp), (i2, d1),..., (ip, dp−1)] ΓbX,Y i

Now by dividing by |X|, we obtain the formula in the statement.

The formula in Theorem 12.17 can be interpreted as follows:

Theorem 12.18. With M = |X|,N = |Y | we have the formula

1 1 law(χ) = 1 − δ + law(A) N 0 N

MN where A ∈ C(T ,MM (C)) is given by A(q) = Gram matrix of the rows of q. 192 TEO BANICA

Proof. According to Theorem 12.17, we have the following formula: Z p 1 X X χ = δ[i1d1,...,ipdp],[i1dp,...,ipdp−1] G MN i1...ip d1...dp Z 1 X X qi1d1 . . . qipdp = dq MN MN qi1dp . . . qipdp−1 T i1...ip d1...dp   Z ! ! 1 X X qi d X qi d X qipdp = 1 1 2 2 ...   dq MN MN qi2d1 qi3d2 qi1dp T i1...ip d1 d2 dp

Consider now the Gram matrix in the statement, A(q)ij =< Ri,Rj >, where R1,...,RM MN are the rows of q ∈ T ' MM×N (T). We have then: Z Z p 1 χ = < Ri1 ,Ri2 >< Ri2 ,Ri3 > . . . < Rip ,Ri1 > G MN TMN 1 Z = A(q)i1i2 A(q)i2i3 ...A(q)ipi1 MN TMN 1 Z = T r(A(q)p)dq MN TMN 1 Z = tr(A(q)p)dq N TMN But this gives the formula in the statement, and we are done. The problem now is that of ﬁnding the good regime, M = f(K),N = g(K),K → ∞, where the measure in Theorem 12.18 converges, after some suitable manipulations. We denote by NC(p) the set of noncrossing partitions of {1, . . . , p}, and for π ∈ P (p) we denote by |π| ∈ {1, . . . , p} the number of blocks. We will need: Proposition 12.19. With M = αK, N = βK, K → ∞ we have: p cp X n o ' # π ∈ NC(p) |π| = r αr−1βp−r Kp−1 r=1

1 2p p−1 In particular, with α = β we have cp ' p+1 p (αK) . Proof. We use the combinatorial formula in Theorem 12.17 above. Our claim is that, with π = ker(i1, . . . , ip), the corresponding contribution to cp is: ( α|π|−1βp−|π|Kp−1 if π ∈ NC(p) C ' π O(Kp−2) if π∈ / NC(p) HADAMARD MATRICES 193

As a ﬁrst observation, since there are M(M − 1) ... (M − |π| + 1) ' M |π| choices for a p multi-index (i1, . . . , ip) ∈ X satisfying ker i = π, we have: n o |π|−1 −1 Cπ ' M N # d1, . . . , dp ∈ Y [dα|α ∈ b] = [dα−1|α ∈ b], ∀b ∈ π

Consider now the partition σ = ker d. The contribution of σ to the above quantity Cπ is then given by ∆(π, σ)N(N − 1) ... (N − |σ| + 1) ' ∆(π, σ)N |σ|, where: ( 1 if |b ∩ c| = |(b − 1) ∩ c|, ∀b ∈ π, ∀c ∈ σ ∆(π, σ) = 0 otherwise

We use now the standard fact that for π, σ ∈ P (p) satisfying ∆(π, σ) = 1 we have:

|π| + |σ| ≤ p + 1

In addition, the equality case happens when π, σ ∈ NC(p) are inverse to each other, p−2 via Kreweras complementation. This shows that for π∈ / NC(p) we have Cπ = O(K ), and that for π ∈ NC(p) we have:

|π|−1 −1 p−|π|−1 Cπ ' M N N = α|π|−1βp−|π|Kp−1

Thus, we have obtained the result.

We denote by D the dilation operation, Dr(law(X)) = law(rX). We have: Theorem 12.20. With M = αK, N = βK, K → ∞ we have: 1 1 µ = 1 − δ0 + D 1 (πα/β) αβK2 αβK2 βK

1 1 In particular with α = β we have µ = 1 − 2 2 δ0 + 2 2 D 1 (π1). α K α K αK Proof. At α = β, this follows from Proposition 12.19. In general now, we have:

cp X ' α|π|−1βp−|π| Kp−1 π∈NC(p) |π| βp X α = α β π∈NC(p) βp Z = xpdπ (x) α α/β 194 TEO BANICA

When α ≥ β, where dπα/β(x) = ϕα/β(x)dx is continuous, we obtain: 1 Z c = (βKx)pϕ (x)dx p αK α/β 1 Z x = xpϕ dx αβK2 α/β βK But this gives the formula in the statement. When α ≤ β the computation is similar, with a Dirac mass as 0 dissapearing and reappearing, and gives the same result.

As a ﬁrst comment, when interchanging α, β we obtain D 1 (πα/β) = D 1 (πβ/α), which βK αK is a consequence of the well-known formula πt−1 = Dt(πt). This latter formula is best understood by using Kreweras complementation, which gives indeed: Z p X |π| x dπt(x) = t π∈NC(p) X = tp+1 t−|π| π∈NC(p) Z p = t (tx) dπt−1 (x) Let us state as well an explicit result, regarding densities: Theorem 12.21. With M = αK, N = βK, K → ∞ we have: 1 1 p4αβK2 − (x − αK − βK)2 µ = 1 − δ + · dx αβK2 0 αβK2 2πx √ 4αK 1 1 x −1 In particular with α = β we have µ = 1 − α2K2 δ0 + α2K2 · 2π . Proof. According to the formula for the density of the free Poisson law,, the density of the continuous part D 1 (πα/β) is indeed given by: βK q 4 α − ( x − 1 − α )2 p β βK β 4αβK2 − (x − αK − βK)2 x = 2π · βK 2πx With α = β now, we obtain the second formula in the statement, and we are done. Observe that at α = β = 1, where M = N = K → ∞, the measure in Theorem 12.21, 1 1 namely µ = 1 − 2 δ0 + 2 D 1 (π1), is supported by [0, 4K]. On the other hand, since K K K the groups ΓM,N are all amenable, the corresponding measures are supported on [0,MN], and so on [0,K2] in the M = N = K situation. The fact that we don’t have a convergence of supports is not surprising, because our convergence is in moments. We have as well the following result, which includes computations from [9]: HADAMARD MATRICES 195

Theorem 12.22. Given two finite abelian groups G, H, with |G| = M, |H| = N, consider the main character χ of the quantum group associated to FG×H . We have then χ 1 1 law = 1 − δ + π N M 0 M t in moments, with M = tN → ∞, where πt is the free Poisson law of parameter t > 0. In addition, this formula holds for any generic fiber of FG×H . Proof. We already know that the second assertion holds, as explained in Theorem 12.21. Regarding now the first assertion, which is from [9], our first claim is that for the representation coming from the parametric matrix FG×H we have the following formula, where M = |G|,N = |H|, and the sets between brackets are sets with repetitions:  i , . . . , i , a , . . . , a ∈ {0,...,M − 1},   1 r 1 p  r 1  b1, . . . , bp ∈ {0,...,N − 1},  cp = # M r+1N [(ix + ay, by), (ix+1 + ay, by+1)|y = 1, . . . , p]   = [(ix + ay, by+1), (ix+1 + ay, by)|y = 1, . . . , p], ∀x

Indeed, by using the general moment formula with K = FG, L = FH , we have: 1 1 1 1 r r r r Z Q 1 1 Q 2 1 Qi1b1 Q 2 1 Qirbr Q 1 r Qir br Q 1 r r 1 X X i1b1 i1b2 p p ipb1 1 1 i1b2 p p ipb1 cp = r 1 1 ... 1 1 ...... r r ... r r (MN) T r Q 1 1 Q 2 1 Q 1 1 Q 2 1 Q r r Q 1 r Q r r Q 1 r 1 r 1 r i b i b ipb ipbp i1b2 i b ipb1 ipbp i1...ip b1...bp 1 2 1 1 1 1 1

K 1 1 K 2 1 Ki1j1 K 2 1 K r r K 1 r K r r K 1 r 1 X i1j1 i1j2 p p ipj1 i1j1 i1j2 ipjp ipj1 pr ...... M K 1 1 K 2 1 K 1 1 K 2 1 K r r K 1 r K r r K 1 r 1 r i1j2 i1j1 ipj1 ipjp i1j2 i1j1 ipj1 ipjp j1 ...jp

L 1 1 L 2 1 La1b1 L 2 1 L r r L 1 r L r r L 1 r 1 X a1b1 a1b2 p p apb1 a1b1 a1b2 apbp apb1 pr ...... dQ N L 1 1 L 2 1 L 1 1 L 2 1 L r r L 1 r L r r L 1 r 1 r a1b2 a1b1 apb1 apbp a1b2 a1b1 apb1 apbp a1...ap

Since we are in the Fourier matrix case, K = FG,L = FH , we can perform the sums over j, a. To be more precise, the last two averages appearing above are respectively:

Y Y x x+1 x+1 x ∆(i) = δ(iy + iy−1, iy + iy−1) x y Y Y x x+1 x+1 x ∆(b) = δ(by + by−1, by + by−1) x y We therefore obtain the following formula for the truncated moments of the main character, where ∆ is the product of Kronecker symbols constructed above: 1 1 1 1 r r r r Z Q 1 1 Q 2 1 Qi1b1 Q 2 1 Qirbr Q 1 r Qir br Q 1 r r 1 X i1b1 i1b2 p p ipb1 1 1 i1b2 p p ipb1 cp = r 1 1 ... 1 1 ...... r r ... r r dQ (MN) T r Q 1 1 Q 2 1 Q 1 1 Q 2 1 Qirbr Q 1 r Qir br Q 1 r ∆(i)=∆(b)=1 i1b2 i1b1 ipb1 ipbp 1 2 i1b1 p 1 ipbp 196 TEO BANICA

Now by integrating with respect to Q ∈ (TG×H )r, we are led to counting the multi- indices i, b satisfying the condition ∆(i) = ∆(b) = 1, along with the following conditions, where the sets between brackets are by deﬁnition sets with repetitions: 1 1 1 1 2 1 2 1 1 1 1 1 2 1 2 1 i1b1 . . . ipbp i1b2 . . . ipb1 = i1b2 . . . ipb1 i1b1 . . . ipbp . . r r r r 1 r 1 r r r r r 1 r 1 r i1b1 . . . ipbp i1b2 . . . ipb1 = i1b2 . . . ipb1 i1b1 . . . ipbp In a more compact notation, the moment formula is therefore as follows: 1 n o cr = # i, b ∆(i) = ∆(b) = 1, [ixbx, ix+1bx ] = [ixbx , ix+1bx], ∀x p (MN)r y y y y+1 y y+1 y y Now observe that the above Kronecker type conditions ∆(i) = ∆(b) = 1 tell us that x x the arrays of indices i = (iy ), b = (by ) must be of the following special form:  1 1    1 1   i1 . . . ip i1 + a1 . . . i1 + ap b1 . . . bp j1 + b1 . . . j1 + bp  ...  =  ...  ,  ...  =  ...  1 r 1 r ir . . . ip ir + a1 . . . ir + ap br . . . bp jr + b1 . . . jr + bp

Here all the new indices ix, jx, ay, by are uniquely determined, up to a choice of i1, j1. x x Now by replacing iy , by with these new indices ix, jx, ay, by, with a MN factor added, which accounts for the choice of i1, j1, we obtain the following formula: r 1 [(ix + ay, jx + by), (ix+1 + ay, jx + by+1)] cp = # i, j, a, b (MN)r+1 = [(ix + ay, jx + by+1), (ix+1 + ay, jx + by)], ∀x

Now observe that we can delete if we want the jx indices, which are irrelevant. Thus, we obtain the announced formula. The continuation is via combinatorics, see [9].

As already mentioned, on several occasions, such computations are expected to have some applications in statistical mechanics, but this remains to be worked out. There are many other possible applications of the complex Hadamard matrices. Let us mention for instance some potential relations with noncommutative geometry, particle physics, and CKM type matrices, in the spirit of the paper of Connes [38]. Some other well-known applications of the Hadamard matrices concern questions in quantum information theory. We refer here to the MUB literature [26], [48], [89], with the remark that the relation of this with the above still remains to be understood. This list, which adds to the previous considerations in this book, is certainly incomplete, the Fourier type matrices being potentially useful a bit everywhere. HADAMARD MATRICES 197

References [1] S. Agaian, Hadamard matrices and their applications, Springer (1985). [2] J. Avan, T. Fonseca, L. Frappat, P. Kulish, E. Ragoucy and G. Rollet, Temperley-Lieb R-matrices from generalized Hadamard matrices, Theor. Math. Phys. 178 (2014), 223–240. [3] J. Backelin, Square multiples n give inﬁnitely many cyclic n-roots, preprint 1989. [4] T. Banica, Compact Kac algebras and commuting squares, J. Funct. Anal. 176 (2000), 80–99. [5] T. Banica, The defect of generalized Fourier matrices, Linear Algebra Appl. 438 (2013), 3667–3688. [6] T. Banica, First order deformations of the Fourier matrix, J. Math. Phys. 55 (2014), 1–22. [7] T. Banica, Counting results for thin Butson matrices, Electron. J. Combin. 21 (2014), 1–14. [8] T. Banica, The glow of Fourier matrices: universality and ﬂuctuations, Oper. Matrices 9 (2015), 457–474. [9] T. Banica, Deformed Fourier models with formal parameters, Studia Math. 239 (2017), 201–224. [10] T. Banica, Complex Hadamard matrices with noncommutative entries, Ann. Funct. Anal. 9 (2018), 354–368. [11] T. Banica and J. Bichon, Random walk questions for linear quantum groups, Int. Math. Res. Not. 24 (2015), 13406–13436. [12] T. Banica, J. Bichon and B. Collins, Quantum permutation groups: a survey, Banach Center Publ. 78 (2007), 13–34. [13] T. Banica, J. Bichon and J.-M. Schlenker, Representations of quantum permutation algebras, J. Funct. Anal. 257 (2009), 2864–2910. [14] T. Banica, B. Collins and J.-M. Schlenker, On orthogonal matrices maximizing the 1-norm, Indiana Univ. Math. J. 59 (2010), 839–856. [15] T. Banica and I. Nechita, Almost Hadamard matrices: the case of arbitrary exponents, Discrete Appl. Math. 161 (2013), 2367–2379. [16] T. Banica and I. Nechita, Flat matrix models for quantum permutation groups, Adv. Appl. Math. 83 (2017), 24–46. [17] T. Banica and I. Nechita, Almost Hadamard matrices with complex entries, Adv. Oper. Theory 3 (2018), 149–189. [18] T. Banica, I. Nechita and J.-M. Schlenker, Analytic aspects of the circulant Hadamard conjecture, Ann. Math. Blaise Pascal 21 (2014), 25–59. [19] T. Banica, I. Nechita and J.-M. Schlenker, Submatrices of Hadamard matrices: complementation results, Electron. J. Linear Algebra 27 (2014), 197–212. [20] T. Banica, I. Nechita and K. Zyczkowski,˙ Almost Hadamard matrices: general theory and examples, Open Syst. Inf. Dyn. 19 (2012), 1–26. [21] T. Banica and R. Nicoara, Quantum groups and Hadamard matrices, Panamer. Math. J. 17 (2007), 1–24. [22] T. Banica, D. Ozteke¨ and L. Pittau, Isolated partial Hadamard matrices, and related topics, Open Syst. Inf. Dyn. 25 (2018), 1–27. [23] T. Banica and A. Skalski, The quantum algebra of partial Hadamard matrices, Linear Algebra Appl. 469 (2015), 364–380. [24] L.D. Baumert, S.W. Golomb and M. Hall, Discovery of an Hadamard matrix of order 92, Bull. Amer. Math. Soc. 68 (1962), 237–238. [25] K. Beauchamp and R. Nicoara, Orthogonal maximal abelian ∗-subalgebras of the 6 × 6 matrices, Linear Algebra Appl. 428 (2008), 1833–1853. [26] I. Bengtsson, W. Bruzda, A.˚ Ericsson, J.-A.˚ Larsson, W. Tadej and K. Zyczkowski,˙ Mutually unbiased bases and Hadamard matrices of order six, J. Math. Phys. 48 (2007), 1–33. 198 TEO BANICA

[27] H. Bercovici and V. Pata, Stable laws and domains of attraction in free probability theory, Ann. of Math. 149 (1999), 1023–1060. [28] J. Bichon, Quotients and Hopf images of a smash coproduct, Tsukuba J. Math. 39 (2015), 285–310. [29] P. Biran, M. Entov and L. Polterovich, Calabi quasimorphisms for the symplectic ball, Commun. Contemp. Math. 6 (2004), 793–802. [30] G. Björck, Functions of modulus 1 on Zn whose Fourier transforms have constant modulus, and cyclic n-roots, NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci. 315 (1990), 131–140. [31] G. Björck and R. Fröberg, A faster way to count the solutions of inhomogeneous systems of algebraic equations, with applications to cyclic n-roots, J. Symbolic Comput. 12 (1991), 329–336. [32] G. Björck and U. Haagerup, All cyclic p-roots of index 3 found by symmetry-preserving calculations, preprint 2008. [33] R. Burstein, Group-type subfactors and Hadamard matrices, Trans. Amer. Math. Soc. 367 (2015), 6783–6807. [34] A.T. Butson, Generalized Hadamard matrices, Proc. Amer. Math. Soc. 13 (1962), 894–898. [35] C.-H. Cho, Holomorphic discs, spin structures, and Floer cohomology of the Clifford torus, Int. Math. Res. Not. 35 (2004), 1803–1843. [36] B. Collins and P. Sniady,´ Integration with respect to the Haar measure on unitary, orthogonal and symplectic groups, Comm. Math. Phys. 264 (2006), 773–795. [37] A. Connes, Une classification des facteurs de type III, Ann. Sci. Ec. Norm. Sup. 6 (1973), 133–252. [38] A. Connes, A unitary invariant in Riemannian geometry, Int. J. Geom. Methods Mod. Phys. 5 (2008), 1215–1242. [39] R. Craigen and H. Kharaghani, On the nonexistence of Hermitian circulant complex Hadamard matrices, Australas. J. Combin. 7 (1993), 225–227. [40] W. de Launey, On the non-existence of generalized weighing matrices, Ars Combin. 17 (1984), 117–132. [41] W. de Launey and J.E. Dawson, An asymptotic result on the existence of generalised Hadamard matrices, J. Combin. Theory Ser. A 65 (1994), 158–163. [42] W. de Launey, D.L. Flannery and K.J. Horadam, Cocyclic Hadamard matrices and difference sets, Discrete Appl. Math. 102 (2000), 47–61. [43] W. de Launey and D.M. Gordon, A comment on the Hadamard conjecture, J. Combin. Theory Ser. A 95 (2001), 180–184. [44] W. de Launey and D.A. Levin, A Fourier-analytic approach to counting partial Hadamard matrices, Cryptogr. Commun. 2 (2010), 307–334. [45] P. Di Francesco, Meander determinants, Comm. Math. Phys. 191 (1998), 543–583. [46] P. Diaconis and M. Shahshahani, On the eigenvalues of random matrices, J. Applied Probab. 31 (1994), 49–62. [47] P. Dit¸˘a,Some results on the parametrization of complex Hadamard matrices, J. Phys. A 37 (2004), 5355–5374. [48] T. Durt, B.-G. Englert, I. Bengtsson and K. Zyczkowski,˙ On mutually unbiased bases, Int. J. Quan- tum Inf. 8 (2010), 535–640. [49] J.-C. Faugère,Finding all the solutions of Cyclic 9 using Gröbnerbasis techniques, Lecture Notes Ser. Comput. 9 (2001), 1–12. [50] P.C. Fishburn and N.J.A. Sloane, The solution to Berlekamp’s switching game, Discrete Math. 74 (1989), 263–290. [51] U. Haagerup, Orthogonal maximal abelian ∗-subalgebras of the n × n matrices and cyclic n-roots, in “Operator algebras and quantum field theory”, International Press (1997), 296–323. HADAMARD MATRICES 199

[52] U. Haagerup, Cyclic p-roots of prime lengths p and related complex Hadamard matrices, preprint 2008. [53] J. Hadamard, Résolutiond’une question relative aux déterminants, Bull. Sci. Math. 2 (1893), 240– 246. [54] M. Hall, Integral matrices A for which AAT = mI, in “Number Theory and Algebra”, Academic Press (1977), 119–134. [55] G. Hiranandani and J.-M. Schlenker, Small circulant complex Hadamard matrices of Butson type, European J. Combin. 51 (2016), 306–314. [56] K.J. Horadam, Hadamard matrices and their applications, Princeton Univ. Press (2007). [57] M. Idel and M.M. Wolf, Sinkhorn normal form for unitary matrices, Linear Algebra Appl. 471 (2015), 76–84. [58] N. Ito, Hadamard Graphs I, Graphs Combin. 1 (1985), 57–64. [59] V.F.R. Jones, Index for subfactors, Invent. Math. 72 (1983), 1–25. [60] V.F.R. Jones, On knot invariants related to some statistical mechanical models, Pacific J. Math. 137 (1989), 311–334. [61] V.F.R. Jones, Planar algebras I, preprint 1999. [62] A. Karabegov, The reconstruction of a unitary matrix from the moduli of its elements and symbols on a finite phase space, preprint 1989. [63] H. Kharaghani and J. Seberry, The excess of complex Hadamard matrices, Graphs Combin. 9 (1993), 47–56. [64] H. Kharaghani and B. Tayfeh-Rezaie, A Hadamard matrix of order 428, J. Combin. Des. 13 (2005), 435–440. [65] C. Koukouvinos, M. Mitrouli and J. Seberry, An algorithm to find formulae and values of minors for Hadamard matrices, Linear Algebra Appl. 330 (2001), 129–147. [66] T.Y. Lam and K.H. Leung, On vanishing sums of roots of unity, J. Algebra 224 (2000), 91–109. [67] V.A. Marchenko and L.A. Pastur, Distribution of eigenvalues in certain sets of random matrices, Mat. Sb. 72 (1967), 507–536. [68] M. Matolcsi, J. Réffyand F. Szöll˝osi,Constructions of complex Hadamard matrices via tiling abelian groups, Open Syst. Inf. Dyn. 14 (2007), 247–263. [69] D. McNulty and S. Weigert, Isolated Hadamard matrices from mutually unbiased product bases, J. Math. Phys. 53 (2012), 1–21. [70] M.T. Mohan, On some p-almost Hadamard matrices, Oper. Matrices 13 (2019), 253–281. [71] F.J. Murray and J. von Neumann, On rings of operators. IV, Ann. of Math. 44 (1943), 716–808. [72] R. Nicoara, A finiteness result for commuting squares of matrix algebras, J. Operator Theory 55 (2006), 295–310. [73] R. Nicoara and J. White, Analytic deformations of group commuting squares and complex Hadamard matrices, J. Funct. Anal. 272 (2017), 3486–3505. [74] A. Ocneanu, Quantum symmetry, differential geometry of finite graphs, and classification of subfactors, Univ. of Tokyo Seminary Notes (1990). [75] R. Paley, On orthogonal matrices, J. Math. Phys. 12 (1933), 311–320. [76] K.-H. Park and H.-Y. Song, Quasi-Hadamard matrices, Proc. ISIT 2010, Austin, TX (2010). [77] M. Petrescu, Existence of continuous families of complex Hadamard matrices of certain prime dimensions and related results, Ph.D. Thesis, UCLA (1997). [78] G. Pólya, Uber¨ eine Aufgabe der Wahrscheinlichkeitsrechnung betreffend die Irrfahrt im Strassen- netz, Math. Ann. 84 (1921), 149–160. [79] S. Popa, Orthogonal pairs of ∗-subalgebras in finite von Neumann algebras, J. Operator Theory 9 (1983), 253–268. 200 TEO BANICA

[80] S. Popa, Classification of amenable subfactors of type II, Acta Math. 172 (1994), 163–255. [81] L.B. Richmond and J. Shallit, Counting abelian squares, Electron. J. Combin. 16 (2009), 1–9. [82] R. Roth and K. Viswanathan, On the hardness of decoding the Gale-Berlekamp code, IEEE Trans. Inform. Theory 54 (2008), 1050–1060. [83] H.J. Ryser, Combinatorial mathematics, Wiley (1963). [84] J. Seberry and M. Yamada, Hadamard matrices, sequences, and block designs, Wiley (1992). [85] J.J. Sylvester, Thoughts on inverse orthogonal matrices, simultaneous sign-successions, and tesse- lated pavements in two or more colours, with applications to Newton’s rule, ornamental tile-work, and the theory of numbers, Phil. Mag. 34 (1867), 461–475. [86] F. Szöll˝osi,Parametrizing complex Hadamard matrices, European J. Combin. 29 (2008), 1219–1234. [87] F. Szöll˝osi,Exotic complex Hadamard matrices and their equivalence, Cryptogr. Commun. 2 (2010), 187–198. [88] W. Tadej and K. Zyczkowski,˙ A concise guide to complex Hadamard matrices, Open Syst. Inf. Dyn. 13 (2006), 133–177. [89] W. Tadej and K. Zyczkowski,˙ Defect of a unitary matrix, Linear Algebra Appl. 429 (2008), 447–481. [90] T. Tao, Fuglede’s conjecture is false in 5 and higher dimensions, Math. Res. Lett. 11 (2004), 251–258. [91] T. Tao and V. Vu, On random ±1 matrices: singularity and determinant, Random Structures Algo- rithms 28 (2006), 1–23. [92] N.H. Temperley and E.H. Lieb, Relations between the “percolation” and “colouring” problem and other graph-theoretical problems associated with regular planar lattices: some exact results for the “percolation” problem, Proc. Roy. Soc. London 322 (1971), 251–280. [93] R.J. Turyn, Character sums and difference sets, Pacific J. Math. 15 (1965), 319–346. [94] E. Verheiden, Integral and rational completions of combinatorial matrices, J. Combin. Theory Ser. A 25 (1978) 267–276. [95] D.V. Voiculescu, K.J. Dykema and A. Nica, Free random variables, AMS (1992). [96] S. Wang, Quantum symmetry groups of finite spaces, Comm. Math. Phys. 195 (1998), 195–211. [97] J. Williamson, Hadamard’s determinant theorem and the sum of four squares, Duke Math. J. 11 (1944), 65–81. [98] A. Winterhof, On the non-existence of generalized Hadamard matrices, J. Statist. Plann. Inference 84 (2000), 337–342. [99] S.L. Woronowicz, Compact matrix pseudogroups, Comm. Math. Phys. 111 (1987), 613–665. [100] S.L. Woronowicz, Tannaka-Krein duality for compact matrix pseudogroups. Twisted SU(N) groups, Invent. Math. 93 (1988), 35–76.

T.B.: Department of Mathematics, University of Cergy-Pontoise, F-95000 Cergy- Pontoise, France. [email protected]